A) \[6.25A,\,\,3.75V\]
B) \[3.00A,\,5V\]
C) \[3.75A,\,3.75V\]
D) \[6.25A,6.25V\]
Correct Answer: B
Solution :
\[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}\] (Series combination) \[{{R}_{eq}}\]of the circuit \[=\frac{5\times 15}{5+15}+\frac{125}{100}=\frac{75}{20}+\frac{5}{4}5\Omega \] We know that \[I=\frac{E}{{{R}_{eq}}}=\frac{20}{5}=4A\] Potential difference across \[{{R}_{1}}\]and \[{{R}_{2}}\] are same (Parallel combination) \[{{I}_{1}}{{R}_{1}}=(4-{{I}_{1}}){{R}_{2}}\] \[\Rightarrow \]\[5{{I}_{1}}=(4-{{I}_{1}})\times 15\Rightarrow {{I}_{1}}=12-3{{I}_{1}}\Rightarrow {{I}_{1}}=3A\] Thus reading of ammeter = 3A Voltage across \[1.25\Omega =I\times R=4\times 1.25=5V\] (Reading of voltmeter)You need to login to perform this action.
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