Explanation
To convert a galvanometer into a voltmeter a high resistance is connected in series with galvanometer.
To shift the balance point on higher length the potential gradient of wire is to be decreased. The same can be obtained by decreasing the current of the main circuit which is possible by increasing the resistance in main circuit of potentiometer wire.
A voltmeter is a high resistance device and is always connected in parallel with the circuit. While an ammeter is a low resistance device and is always connected in series with the circuit.So to use voltmeter in place of ammeter a high resistance must be connected in series with the circuit.
Slope of the graph will give us reciprocal of resistance.Here resistance at temp.T1 is greater than that at T2. Since resistance of metallic wire is more at higher temp. then at lower temperature, hence T1 > T2.
Here all the three resistances are in parallel between points A and B. Their equivalent resistance between A and B = R/3.
Pot. diff across first 10Ω = 10 × 0.5 = 5V
Current through 20 Ω = 5/20 = 0.25 A
Current through middle 10 Ω = 5/10 = 0.5 A
Effective e.m.f. of circuit =10 – 3 = 7V
Total resistance of circuit = 2 + 5 + 3 + 4 = 14Ω
Current, I = 7/14 = 0.5 A
Pot. diff. between A and D = 0.5 × 10 = 5V
Potential at D = 10 – 5 = 5V.
∴Potentialat E = 5 – 3 = 2V.
Hence E can not be at zero potential, as there is potential drop at E.
current, I = 2/20 = 1/10 amp.
Potential difference across 2Ω resistance from left side cell is 10 V. Therefore, potential difference across 2Ω resistance from right side cell should also be 10 V. This is possible only if E = 10 V as the current passing through 1Ω resistance is zero.
On short circuiting E2, the resistance R2 will reduce to zero as R2 is in parallel to E2. So the current in the circuit will be E1/R1.
When n resistances are in parallel, then equivalent resistance is R = r/n or r = nR. When n resistance are in series, then equivalent resistance, R’ = nr = n (nR) = n2 R
Brown → 1
Yellow → 4
Green → 105
Gold → (± 5%)
R = 14 ×105 ± 5%
= (1.4 ± 0.05) M Ω
Charge is equal to area under the curve of current and time graph.
For 1st rectangle, q1 = l × b =2
For 2nd rectangle, q2 = l × b =2
For 3rd triangle, q 3 =1/2 l b =2
Hence, ratio is 1:1:1.
Ammeter is always connected in series and voltmeter is connected in parallel.
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