Current Electricity - Result Question 99
104. The resistance of the four arms P, Q, R and S in a Wheatstone’s bridge are $10 ohm, 30 ohm, 30$ ohm and $90 ohm$, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is $50 ohm$, the current drawn from the cell will be
[2013]
(a) $0.2 A$
(b) $0.1 A$
(c) $2.0 A$
(d) $1.0 A$
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Answer:
Correct Answer: 104. (a)
Solution:
- (a) Given: $V=7 V$
$ \begin{aligned} & R _{\text{eq }}=\frac{40 \times 120}{40+120} \Omega \\ & I=\frac{V}{R}=\frac{7}{5+\frac{40 \times 120}{40+120}} \\ & =\frac{7}{5+30}=\frac{1}{5}=0.2 A . \end{aligned} $
