Physics - Current Electricity - Quiz-8

Two wires of resistance R₁ and R₂ have temperature coefficient of resistance $α_{1} and α_{2}$ , respectively. These are joined in series. The effective temperature coefficient of resistance is :

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$\frac{α_{1} + α_{2}}{2}$
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$\sqrt{α_{1} α_{2}}$
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$\frac{α_{1} R_{1} + α_{2} R_{2}}{R_{1} + R_{2}}$
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$\frac{\sqrt{R_{1} R_{2} α_{1} α_{2}}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}$

Two cells of equal e.m.f. and of internal resistances, r_1, and $r_{2} (r_{1} > r_{2})$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be :

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$r_{1} + r_{2}$
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$r_{1} - r_{2}$
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$\frac{r_{1} + r_{2}}{2}$
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$\frac{r_{1} - r_{2}}{2}$

When connected across the terminals of a cell, a voltmeter measures 5V and a connected ammeter measures 10 A of current. A resistance of 2 ohms is connected across the terminals of the cell. The current flowing through this resistance will be :

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2.5 A
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2.0 A
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5.0 A
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7.5 A

In the circuit shown here, E₁ = E₂= E₃ = 2 V and R₁ = R₂ = 4 ohms. The current flowing between points A and B through battery E₂ is

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Zero
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2 amp from A to B
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2 amp from B to A
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None of the above

In the circuit shown below E₁ = 4.0 V, R₁= 2 Ω, E₂ = 6.0 V, R₂ = 4 Ω and R₃ = 2 Ω. The current I₁ is :

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1.6 A
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1.8 A
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1.25 A
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1.0 A

The potential difference across 8 ohms resistance is 48 volts as shown in the figure. The value of potential difference across X and Y points will be :

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160 volt
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128 volt
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80 volt
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62 volt

Two resistances R₁ and R₂ are made of different materials. The temperature coefficient of the material of R₁ is α and of the material of R₂ is –β. The resistance of the series combination of R₁ and R₂ will not change with temperature, if R₁/ R₂ equals :

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$\frac{α}{β}$
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$\frac{α + β}{α - β}$
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$\frac{α^{2} + β^{2}}{α β}$
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$\frac{β}{α}$

An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^{7}$ electrons and the same number of singly-charged positive ions per cm³. The electrons are moving at 0.4 m/s. The current density from anode to cathode is $4 μ A / m^{2}$ . The velocity of positive ions moving towards cathode is :

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0.4 m/s
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16 m/s
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Zero
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0.1 m/s

A wire of resistance 10 Ω is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a Battery of 3 V and internal resistance 1 Ω as shown in the figure. The currents in the two parts of the circle are

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$\frac{6}{23} A and \frac{18}{23} A$
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$\frac{5}{26} A and \frac{15}{26} A$
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$\frac{4}{25} A and \frac{12}{25} A$
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$\frac{3}{25} A and \frac{9}{25} A$

In the given circuit, it is observed that the current I is independent of the value of the resistance R₆. Then the resistance values must satisfy

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$R_{1} R_{2} R_{5} = R_{3} R_{4} R_{6}$
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$\frac{1}{R_{5}} + \frac{1}{R_{6}} = \frac{1}{R_{1} + R_{2}} + \frac{1}{R_{3} + R_{4}}$
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$R_{1} R_{4} = R_{2} R_{3}$
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$R_{1} R_{3} = R_{2} R_{4} = R_{5} R_{6}$

In the given circuit, with a steady current, the potential drop across the capacitor must be :

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V
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V / 2
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V / 3
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2V / 3

A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to current, the temperature of the wire is raised by ΔT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross–section but of length 2 L. The temperature of the wire is raised by the same amount ΔT in the same time t. The value of N is-

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8
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9
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4
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6

What is the equivalent resistance between points A and B of the network?

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$\frac{57}{7} Ω$
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8 Ω
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6 Ω
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$\frac{57}{5} Ω$

The effective resistance between points P and Q of the electrical circuit shown in the figure is:

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$2 R r / (R + r)$
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$8 R (R + r) / (3 R + r)$
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$2 r + 4 R$
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$5 R / 2 + 2 r$

In the circuit element given here, if the potential at point B, V_B = 0, then the potentials of A and D are given as

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(1) $V_{A} = - 1 .5 V, V_{D} = + 2 V$
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(2) $V_{A} = + 1 .5 V, V_{D} = + 2 V$
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(3) $V_{A} = + 1 .5 V, V_{D} = + 0 .5 V$
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(4) $V_{A} = + 1 .5 V, V_{D} = - 0 .5 V$

The current in a conductor varies with time t as $I = 2 t + 3 t^{2}$ where I is in ampere and t in seconds. The electric charge flowing through a section of the conductor during t = 2 sec to t = 3 sec is :

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10 C
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24 C
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33 C
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44 C

In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled

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x
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x/4
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4x
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2x

Seven resistances are connected as shown in the figure. The equivalent resistance between A and B is

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3 Ω
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4 Ω
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4.5 Ω
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5 Ω

A battery of internal resistance 4Ω is connected to the network of resistances as shown. In order to give the maximum power to the network, the value of R (in Ω) should be :

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4/9
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8/9
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2
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18

In the circuit shown here, the readings of the ammeter and voltmeter are

(1) 6 A, 60 V

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(3) 6/11 , 60/11 V
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(2) 0.6 A, 6 V
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3
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(4) 11/6 A, 11/60 V

A wire of resistor R is bent into a circular ring of radius r. Equivalent resistance between two points X and Y on its circumference, when angle XOY is α, can be given by

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$\frac{R α}{4 π^{2}} (2 π - α)$
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$\frac{R}{2 π} (2 π - α)$
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R (2π – α)
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$\frac{4 π}{R α} (2 π - α)$

As the switch S is closed in the circuit shown in the figure, the current passed through it is :

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4.5 A
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6.0 A
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3.0 A
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Zero

In the following circuit, a 10 m long potentiometer wire with resistance 1.2 ohm/m, a resistance R₁ and an accumulator of emf 2 V are connected in series. When the emf of the thermocouple is 2.4 mV then the deflection in the galvanometer is zero. The current supplied by the accumulator will be :

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4 × 10^–4 A
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8 × 10^–4 A
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4 × 10^–3 A
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8 × 10^–3 A

In the following circuit, bulb rated as 1.5 V, 0.45 W. If bulbs glows with full intensity then what will be the equivalent resistance between X and Y

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0.45 Ω
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1 Ω
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3 Ω
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5 Ω

Consider the circuits shown in the figure. Both the circuits are taking same current from battery but current through R in the second circuit is $\frac{1}{10}$ th of current through R in the first circuit. If R is 11 Ω, the value of R₁

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9.9 Ω
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11 Ω
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8.8 Ω
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7.7 Ω

In order to quadruple the resistance of a uniform wire, a part of its length was uniformly stretched till the final length of the entire wire was 1.5 times the original length, the part of the wire was fraction equal to

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1 / 8
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1 / 6
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1 / 10
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1 / 4

In the circuit shown in figure reading of voltmeter is V₁ when only S₁ is closed, reading of voltmeter is V₂ when only S₂ is closed and reading of voltmeter is V₃ when both S₁ and S₂ are closed. Then

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V₃ > V₂ > V₁
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V₂ > V₁ > V₃
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V₃ > V₁ > V₂
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V₁ > V₂ > V₃

Current through wire XY of circuit shown is :

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1 A
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4 A
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2 A
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3 A

12 cells each having the same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two similar cells which are in series. Current is 3 A when cells and battery aid each other and is 2 A when cells and battery oppose each other. The number of cells wrongly connected is

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4
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1
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3
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2

Following figure shows cross-sections through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor B will fit snugly within conductor A, and conductor C will fit snugly within conductor B. Relationship between their end to end resistance is

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R_A = R_B = R_C
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R_A > R_B > R_C
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R_A < R_B < R
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Information is not sufficient

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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