Question

The resistances of the four arms $\text{P, Q, R}$ and $\text{S}$ , in a Wheatstone's bridge, are $10\Omega ,30\Omega ,30\Omega \text{and}90\Omega$ respectively. The e.m.f. and internal resistance of the cell are $7\text{V}\text{and}5\Omega$ respectively. If the galvanometer resistance is $50\Omega ,$ the current drawn from the cell will be -

• A. $2.0\text{A}$
• B. $1.0\text{A}$
• C. $0.2\text{A}$
• D. $0.1\text{A}$

Answer 1

The correct option is C $0.2\text{A}$




Since it is a balanced Wheatstone bridge,

For resistance on the left side of the galvanometer,
$R_{S1}=10+30=40\Omega$

For resistance on the right side of the galvanometer,
$R_{S2}=30+90=130\Omega$

Now, $R_{s1}\text{and}{R}_{s2}$ are in parallel, the effective resistance of the Wheatstone's network is,

$R_{eff}=\frac{\left(40\right)\left(120\right)}{40+120}=30\Omega$

Current drawn $I=\frac{E}{R_{eff}+r}=\frac{\text{7}}{(30+5)}=\frac{1}{5}\text{A}=0.2\text{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.