Question

The resistances of the four arms P, Q, R, and S in a Wheatstone’s bridge are $10\Omega$, $30\Omega$, $30\Omega$ and $90\Omega$, respectively. The e.m.f. and the internal resistance of the cell are $7V$ and $5\Omega$ respectively. If the resistance of the galvanometer is $50\Omega$, the current drawn from the cell will be
(a). $1.0A$
(b). $0.2A$
(c). $0.1A$
(d). $2.0A$

Answer 1

- Hint: Draw the circuit diagram for the question, apply the principle of Wheatstone bridge, resolve the circuit and find its equivalent resistance to find the value of current drawn.
Formulae used: Formulae for finding the equivalent resistance in a circuit:

& \text{For series }\Rightarrow \text{ }{{\text{R}}_{eq}}={{\text{R}}_{1}}+{{\text{R}}_{2}} \\\ & \text{For parallel }\Rightarrow \text{ }{{\text{R}}_{eq}}=\dfrac{{{\text{R}}_{1}}{{\text{R}}_{2}}}{{{\text{R}}_{1}}+{{\text{R}}_{2}}} \\\ \end{aligned}$$ Formula for Ohm’s law: $V=IR$ _Complete step-by-step solution_ - Wheatstone bridge is an instrument which is used to measure the unknown resistance connected in a circuit. It consists of four resistors of which two resistors are known resistors, one resistor which can be varied and the unknown resistor. It also consists of a galvanometer. Refer to the connections from the figure. It works on the principle that when the value of resistors follows the relation $\dfrac{P}{Q}=\dfrac{R}{S}$ the voltages at point b and d become the same and the current in the galvanometer G becomes zero. In this question we have $\begin{aligned} & P=10\Omega \\\ & Q=30\Omega \\\ & R=30\Omega \\\ & S=90\Omega \\\ \end{aligned}$ Therefore, $$\dfrac{P}{Q}=\dfrac{10}{30}=\dfrac{R}{S}=\dfrac{30}{90}=\dfrac{1}{3}$$ Thus, no current flows in the galvanometer. And the effective circuit looks like this. We know, $$\begin{aligned} & \text{For series }\Rightarrow \text{ }{{\text{R}}_{eq}}={{\text{R}}_{1}}+{{\text{R}}_{2}} \\\ & \text{For parallel }\Rightarrow \text{ }{{\text{R}}_{eq}}=\dfrac{{{\text{R}}_{1}}{{\text{R}}_{2}}}{{{\text{R}}_{1}}+{{\text{R}}_{2}}} \\\ \end{aligned}$$ Equivalent resistance of the circuit is calculated as follows:  $$\begin{aligned} & {{R}_{eq\\_up}}=P+Q=40\Omega \\\ & {{R}_{eq\\_down}}=R+S=120\Omega \\\ & {{R}_{eq}}=\dfrac{{{R}_{eq\\_up}}{{R}_{eq\\_down}}}{{{R}_{eq\\_up}}+{{R}_{eq\\_down}}}=\dfrac{40\times 120}{40+120}=\dfrac{4800}{160}=30\Omega \\\ \end{aligned}$$ From the relation of voltage, current and resistance in a circuit $V=IR$ The current can be calculated as $I=\dfrac{V}{{{R}_{battery}}+{{R}_{eq}}}=\dfrac{7}{30+5}=\dfrac{7}{35}=0.2A$ _Therefore, the answer to this question is option B. $0.2 A$._ Note: The value of the resistance in the galvanometer is redundant. Do not use that value and complicate the question. Since, in a Wheatstone’s bridge no current flows through the branch with the galvanometer there is no voltage difference because of the resistance in the galvanometer.