Current Electricity - Result Question 45
47. In the circuit shown in the figure, if potential at point $A$ is taken to be zero, the potential at point $B$ is
[2011M]
(a) $-1 V$
(b) $+2 V$
(c) $-2 V$
(d) $+1 V$
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Answer:
Correct Answer: 47. (d)
Solution:
- (d) Current from $D$ to $C=1 A$
$ \begin{aligned} & \therefore V_D-V_C=2 \times 1=2 V \\ & V_A=0 \quad \therefore V_C=1 V, \therefore V_D-V_C=2 \\ & \Rightarrow V_D-1=2 \quad \therefore V_D=3 V \\ & \therefore V_D-V_B=2 \quad \therefore 3-V_B=2 \therefore V_B=1 V \end{aligned} $
(c) $I=\frac{P}{V}=\frac{100 \times 10^{3}}{125} A=\frac{10^{5}}{60} A$
E.C.E. $=0.367 \times 10^{-6} kg C-1$
Charge per minute $=(I \times 60) C$
$=\frac{10^{5} \times 60}{125} C=\frac{6 \times 10^{6}}{125} C$
$\therefore$ Mass liberated, $=\frac{6 \times 10^{6}}{125} \times 0.367 \times 10^{-6}$
$=\frac{6 \times 1000 \times 0.367 \times 10^{-3}}{125}$
$=17.616 \times 10^{-3} kg$
