A) 60V
B) 34V
C) 42 V
D) 56 V
Correct Answer: C
Solution :
\[20\,\Omega \] voltmeter is used in parallel with \[30\,\Omega \]. Hence their effective resistance \[=\frac{30\,\times 20}{30+20}\,=12\Omega \] Now Current through battery \[=\frac{84}{12+12}\,=\frac{84}{24}A\] Hence potential difference across \[12\,\Omega \] resistance \[=iR=\frac{84}{24}\times 12\,=42\] Volt Hence reading of volt meter = 42 voltYou need to login to perform this action.
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