In a series C R circuit shown in the figure, the applied voltage is 10 V and the voltage across capacitor is found to be 8 V. If the phase difference between current and the applied voltage will be tan -1((p/q)), then find p q.

Question

In a series C R circuit shown in the figure, the applied voltage is 10 V and the voltage across capacitor is found to be 8 V. If the phase difference between current and the applied voltage will be tan -1((p/q)), then find p q. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

VR2+VC2=100 VR2+64=100 ⇒ VR=6 V tan φ=(VC/VR)=(8/6)=(4/3)

Q. In a series $CR$ circuit shown in the figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . If the phase difference between current and the applied voltage will be $tan^{- 1} (\frac{p}{q})$ , then find $pq$ .

$V_{R}^{2} + V_{C}^{2} = 100$
$V_{R}^{2} + 64 = 100$
$\Rightarrow V_{R} = 6 V$
$tan ϕ = \frac{V _{C}}{V _{R}} = \frac{8}{6} = \frac{4}{3}$

Q. In a series CR circuit shown in the figure, the applied voltage is 10V and the voltage across capacitor is found to be 8V. If the phase difference between current and the applied voltage will be tan−1(qp​), then find pq.

VR2​+VC2​=100 VR2​+64=100 ⇒VR​=6V tanϕ=VR​VC​​=68​=34​

Q. In a series $CR$ circuit shown in the figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . If the phase difference between current and the applied voltage will be $tan^{- 1} (\frac{p}{q})$ , then find $pq$ .

$V_{R}^{2} + V_{C}^{2} = 100$
$V_{R}^{2} + 64 = 100$
$\Rightarrow V_{R} = 6 V$
$tan ϕ = \frac{V _{C}}{V _{R}} = \frac{8}{6} = \frac{4}{3}$