
- A$8.31$
- B$6.82$
- C$4.92$
- ✓$2$

so effective resistance is $12$ $\Omega$
which is in parallel with $3$ $\Omega$,
so $\frac{1}{R} = \frac{1}{3} + \frac{1}{{12}} = \frac{{15}}{{36}}$
$ \Rightarrow $ $R = \frac{{36}}{{15}}$
$I = \frac{V}{R} = \frac{{4.8 \times 15}}{{36}} = 2\,A$
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$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.