MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equivalent Resistance in Series Circuit

Three resista...

Question

Three resistances are joined together to form a letter $Y$ as shown in the figure.If the potentials of the terminals $A, B$ and $C are V_{1}, V_{2} and V_{3}$ respectively, then determine the potential of the node $O$

A

[\frac{V_{1}}{R_{1}^{2}} + \frac{V_{2}}{R_{2}^{2}} + \frac{V_{3}}{R_{3}^{2}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 2}

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

B

[\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}

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

C

[\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] [R_{1} + R_{2} + R_{3}]

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

D

[\frac{V_{1}}{R_{1}^{2}} + \frac{V_{2}}{R_{2}^{2}} + \frac{V_{3}}{R_{3}^{2}}] [R_{1}^{2} + R_{2}^{2} + R_{3}^{2}]

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

Open in App

Solution

The correct option is B $[\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}$
From the figure,
Applying the junction rule to $O$ , we get

$- I_{1} - I_{2} - I_{3} = 0 \Rightarrow I_{1} + I_{2} + I_{3} = 0$

Now if $V_{0}$ is the potential at point $O$ , then

$I_{1} = \frac{(V_{0} - V_{1})}{R_{1}}; I_{2} = \frac{(V_{0} - V_{2})}{R_{2}}$ and $I_{3} = \frac{(V_{0} - V_{3})}{R_{3}}$ ;

So substituting these values of $I_{1}, I_{2}$ and $I_{3}$ in eq. (i), we get

$V_{0} [\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}] - [\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] = 0$

$V_{0} = [\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A network of resistances is constructed with

R_{1}

R_{2}

as shown in the figure. The potential at the points 1, 2, 3 … n are

V_{1}, V_{2}, V_{3}, \dots V_{n}

respectively, each having a potential k times smaller than the previous one.

The current that passes through the resistance

R_{2}

V_{α}

Q. A network of resistances is constructed with

R_{1}

R_{2}

as shown in the figure. The potential at the points 1, 2, 3 … n are

V_{1}, V_{2}, V_{3}, \dots V_{n}

respectively, each having a potential k times smaller than the previous one.

\frac{R_{2}}{R_{3}}

Q. Two resistances

R_{1}

R_{2}

provide series to parallel equivalents as

\frac{n}{1}

. Then the correct relationship is

Q. In the circuit shown below, the key

K

t = 0

. The current through the battery is

Q. A uniformly charged solid sphere or radius

R

V_{0}

(measured with respect to

\infty

) on its surface. For this sphere, the equipotential surfaces with potentials

\frac{3 V_{0}}{2}, \frac{{5 V}_{0}}{4}, \frac{3 V_{0}}{4}, \frac{V_{0}}{4}

R_{1}, R_{2}, R_{3}

R_{4}

respectively. Then

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equivalent Resistance in Series Circuit

PHYSICS

Watch in App

explore_more_icon

Explore more

Equivalent Resistance in Series Circuit

Standard IX Physics

Join BYJU'S Learning Program

CrossIcon