# (a) An ac circuit as shown in the figure has an inductor of inductance L and a resistor of resistance R connected in series. Using the phasor diagram, explain why the voltage in the circuit will lead the current in phase.(b) The potential difference across the resistor is 160 V and that across the inductor is 120 V. Find the effective value of the applied voltage. If the effective current in the circuit be 1·0 A, calculate the total impedance of the circuit.(c) What will be the potential difference in the circuit when direct current is passed through the circuit?

(a) In an inductor voltage leads current by $\frac{\pi }{2}$ and in resistor both voltage and current are in the same phase, the resultant of $e_{L}$ AND $e_{R}$ that is $e$ in the diagram will lead by phase $\phi .$ So the resultant current will also lead by a phase $\phi$

(b) Given;-

$current\; i=1.0\; A$

$\because V_{R}=160\; V\; \; \; \; \because V_{L}=120\; V$

$R=\frac{V_{R}}{I}=\frac{160}{1}=160\Omega\; \; \; \; \; \; X_{L}=\frac{V_{L}}{I}=\frac{120}{1}=120\Omega$

$V_{net}=\sqrt{V^{2}_{R}+V^{2}_{L}}=\sqrt{(160)^{2}+(120)^{2}}=\sqrt{25600+1440}$

$V_{net}=200$

Now, calculating the impedance (Z) of the circuit.

$Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(160)^{2}+(120)^{2}}=200\Omega$

(c) When direct current is passed through the circuit the potential difference across the inductor is zero.

The potential difference of the circuit = potential difference across the resistor

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