For the given figure, choose the correct option:

V_{a} - V_{f} = 110 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V_{b} - V_{c} = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

V_{a} - V_{b} = 60 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V_{c} - V_{d} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $V_{b} - V_{c} = 0$
Apply KVL;

$q_{1} = \frac{220}{3} μ C$

$q_{2} = - \frac{110}{3} μ C$

$V_{a b} = V_{a} - V_{b} =$ $= \frac{220}{3} V$

$V_{a} - V_{f} =$ $= \frac{220}{3}$ $- \frac{110}{3}$ $- \frac{110}{6}$ $= \frac{110}{6} V$ [KVL]

$V_{b} - V_{c} = 0$

$V_{c} - V_{d} =$ $= \frac{110}{3} V$

Suggest Corrections

Similar questions

Q. Choose the correct relation regarding potential. Here A, B, C and D all are at equal distance from point O, figure. Then ( Assume the dipole to be short)

Choose the correct relation regarding potential of an electric dipole. Here A, B, C and D all are at equal distance from point O which is much larger than the dimensions of the dipole. Then

Q. If

V_{A} - V_{B} = 6 V

, then

V_{C} - V_{D}

A, B and C are voltmeters of resistance R,1.5 R and 3R respectively as shown in the figure.When some potential difference is applied between X and Y, the voltmeter readings are $V_{A}, V_{B}$ and $V_{C}$ respectively.Then :

Q. Compare the speed of the three particles

A, B

and

C

as per the distance-time graph shown below.

Join BYJU'S Learning Program

For the given figure, choose the correct option:

The correct option is B Vb−Vc=0Apply KVL;q1=2203μCq2=−1103μCVab=Va−Vb==2203VVa−Vf==2203−1103−1106=1106V [KVL]Vb−Vc=0 Vc−Vd==1103V