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A circuit when connected to an AC source of $12 \mathrm{~V}$ gives a current of $0.2 \mathrm{~A}$. The same circuit when connected to a DC source of $12 \mathrm{~V}$, gives a current of $0.4 \mathrm{~A}$. The circuit is
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The correct answer is:
series LR
When the circuit is connected to AC source, Voltage, $\mathrm{V}=12 \mathrm{~V}$
Current, $\mathrm{I}=0.2 \mathrm{~A}$
$\Rightarrow$ Impedance $\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.2}=60 \Omega$
When it is connected to DC source,
Voltage, $\mathrm{V}=12 \mathrm{~V}$
Current, $\mathrm{I}=0.4 \mathrm{~A}$
$\Rightarrow$ Resistance, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.4}=30 \Omega$
As in case of DC supply, the capacitor act as an open circuit and no current flows through the circuit. So the given circuit will not have capacitor in series combination. Therefore the circuit should be a series LR circuit.
Current, $\mathrm{I}=0.2 \mathrm{~A}$
$\Rightarrow$ Impedance $\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.2}=60 \Omega$
When it is connected to DC source,
Voltage, $\mathrm{V}=12 \mathrm{~V}$
Current, $\mathrm{I}=0.4 \mathrm{~A}$
$\Rightarrow$ Resistance, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.4}=30 \Omega$
As in case of DC supply, the capacitor act as an open circuit and no current flows through the circuit. So the given circuit will not have capacitor in series combination. Therefore the circuit should be a series LR circuit.
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