The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10 \Omega, 30 \Omega, 30…

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 emf and internal resistance of the cell are $7 \mathrm{~V}$ and $5 \Omega$ respectively. If the galvanometer resistance is $50 \Omega$, the current drawn from the cell will be
  1. $1.0 \mathrm{~A}$
  2. $0.2 \mathrm{~A}$
  3. $0.1 \mathrm{~A}$
  4. $2.0 \mathrm{~A}$

Solution

Effective resistance,
$R_{\text {eff }}=\frac{40 \times 120}{120+40}=\frac{4800}{160}=30 \Omega$
$\therefore$ Current $=\frac{7}{(30+5)}=\frac{7}{35}=0.2 \mathrm{~A}$
$\left[\because i=\frac{E}{R+r}\right]$

Asked in: NEET 2013 (All India)

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