A resistor R and an inductor L are connected in series to a source of voltage V=V_0 \sin \omega t. The voltage is found to lead current in phase by \frac{\pi}{4}. If the inductor is replaced by a capacitor C, the voltage lags behind current in phase by \frac{\pi}{4}. When L, C and R are connected in series with the same source, Find the : (i) average power dissipated and (ii) instantaneous current in the circuit

CBSE Class 12 Physics 2020 Outside Delhi Set 1 Paper

Question : 23

Total: 37

A resistor R and an inductor L are connected in series to a source of voltage V=V0sin‌ωt. The voltage is found to lead current in phase by ‌

.
If the inductor is replaced by a capacitor C, the voltage lags behind current in phase by ‌

. When
L,C and R are connected in series with the same source,
Find the :
(i) average power dissipated and
(ii) instantaneous current in the circuit

Solution:

For LR circuit
‌

=tan‌45∘=1
So, ‌‌XL=R
For CR circuit
‌

=tan‌45∘=1
So, ‌‌Xc=R
For LCR circuit
When L, C and R are connected in series, then actually three R are connected in series. Equivalent impedance =3R. Thus Circuit is resistive.
‌ and ‌‌‌ V‌=V0sin‌ωt
I‌=I0sin‌ωt
(i) Average power dissipation P=VI
P=V0sin‌ωt×I0sin‌ωt
Over full cycle
PAVG=‌

√2

×‌

√2

(ii) Instantaneous current I=I0sin‌ωt

Go to Question:

Prev Question

Next Question