Explanation
The net resistance between X and Y is
Step 1: Calculation of resistance of both wires
(1)&(1) points same potential &(2)&(2) points are same potential. So both the corner Resistance shorted.
Most above with has no resistance. So all are shorted.
I=21=2A
Resistors R1 and R2 have an equivalent resistance of 6 ohms when connected in the circuit shown below. The resistance of R1 could be:
Given that,
The dimension of the block is 1cm, 3cm and 3cm.
The relation of the resistance of a conductor is
R=ρlA....(I)
Where,
ρ = resistivity of the material of the conductor
A= area of the cross section of the conductor
l= length of the conductor
Now, the resistance is maximum
Rmax
{{R}_{\max }}=\dfrac{3\rho }{2}...(II)
Now, the resistance is minimum
{{R}_{\min }}=\rho \dfrac{1}{2\times 3}
{{R}_{\min }}=\dfrac{\rho }{6}...(III)
Now, from equation (II) and (III)
The ratio of maximum resistance and minimum resistance is
\dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{\frac{3\rho }{2}}{\dfrac{\rho }{6}}
\dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{3}{2}\times \dfrac{6}{1}
\dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{9}{1}
{{R}_{\max }}:{{R}_{\min }}=9:1
Hence, the ratio is 9:1
Resistance in wire, R=\rho \dfrac{L}{A}=\dfrac{\rho L}{\pi {{r}^{2}}}
{{R}_{1}}=\dfrac{\rho {{L}_{1}}}{\pi r_{1}^{2}}=\dfrac{\rho {{L}_{1}}}{\pi {{(2r)}^{2}}}=\dfrac{1}{4}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\,\Omega
\Rightarrow \dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\times 4=136\ \Omega \ ......\ (1)
L_2=L_1/2 r_2=r
{{R}_{2}}=\rho \dfrac{{{L}_{2}}}{{{A}_{2}}}=\dfrac{\rho \dfrac{{{L}_{1}}}{2}}{\pi r_{2}^{2}}=\dfrac{1}{2}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=\dfrac{1}{2}\times 136=\ 68\,\Omega
Resistance of B is 68\,\Omega .
\large\textbf{Step -1: Calculate current For Electric bulbs}
Given,
Power of heater P_h= 1500W,
Power of bulbs,P_b = 100W,
We know,
Power, P = VI
The current drawn by the heater,
Or Electrical\;Current,\;I\; = \dfrac{P_h}{V} = \dfrac{{1500}}{{220}} = 6.81\;A
The mains power supply of a house is through a 20 A fuse
Maximum Electric current, which can pass through bulbs = 20- 6.81 =13.19 A
\large\textbf{Step -2:Calculate number of bulbs connected in parallel}
Let number of bulb = n
and Current withdraw by a single bulb, I_b
Electrical\;Current,\;{I_b}\; = \dfrac{P_b}{V} = \dfrac{{100}}{{220}} = 0.45\;A
Maximum number of bulbs, n= 13.19/0.45 = 29
Hence maximum, 29 bulbs can be connected in parallel with heater.
R1 = 24 ± 0.5
R2 = 8 ± 0.3
In series R1 + R2 = 32 ± 0.8
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