In the circuits shown below, the readings of the voltmeters and the ammeters will be
A$\mathrm{V}_{2}>\mathrm{V}_{1}$ and $\mathrm{I}_{1}=\mathrm{I}_{2}$
B$\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{I}_{1}>\mathrm{I}_{2}$
C$\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{I}_{1}=\mathrm{I}_{2}$
D$\mathrm{V}_{2}>\mathrm{V}_{1}$ and $\mathrm{I}_{1}>\mathrm{I}_{2}$
NEET 2019, Medium
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C$\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{I}_{1}=\mathrm{I}_{2}$
c $10 \Omega$ is in series with ideal voltmeter. Therefore it will not affect the circuit
$\mathrm{I}_{1}=\frac{10}{10}=1 \;\mathrm{A}$
$\mathrm{I}_{2}=\frac{10}{10}=1\; \mathrm{A}$
$V_1=10\;V$
$V_{2}=10\;V$
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Consider a block of conducting material ofresistivity '$\rho$' shown in the figure. Current '$I$' enters at '$A$' and leaves from '$D$'. We apply superp osition principle to find voltage '$\Delta V$ ' developed between '$B$' and '$C$'. The calculation is done in the following steps:
$(i)$ Take current '$I$' entering from '$A$' and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculatefield $E(r)$ at distance '$r$' from $A$ by using Ohm's law $E = \rho j$, where j is the current per unit area at '$r$'.
(iii) From the '$r$' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
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