Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Question

In case of the circuit arrangement shown below, the equivalent resistance between A and B is

$10 Ω$
$2.5 Ω$
$\frac{40}{3} Ω$
None of these

A

2.5 Ω

B

None of these

C

10 Ω

D

\frac{40}{3} Ω

Open in App

Solution

Verified by Toppr

There is no resistance in RQ so R and Q are at the same potential.
therefore PQ and PR are at in the parallel so resistance between PQ $= \frac{10 * 10}{10 + 10} = 5$ and that is in series with two 10 $Ω$ resistance so total resistance $= 10 + 10 + 5 Ω = 25 Ω$

Was this answer helpful?

0

Similar Questions

Q1

In case of the circuit arrangement shown below, the equivalent resistance between

A

and

B

is:

View Solution

Q2

Four equal resistors, each of resistance

10 Ω

are connected as shown in the adjoining circuit diagram. Then the equivalent resistance between points A and B is

View Solution

Q3

Four resistances

10 Ω, 5 Ω, 7 Ω

and

3 Ω

are connected so that they form the sides of a rectangle AB, BC, CD and DA respectively. Another resistance of

10 Ω

is connected across the diagonal AC. The equivalent resistance between A and B is

View Solution

Q4

Four resistances

10 Ω, 5 Ω, 7 Ω

and

3 Ω

are connected so that they form the sides of a rectangle

A B, B C, C D

and

D A

respectively. Another resistance of

10 Ω

is connected across the diagonal

A C

. The equivalent resistance between

A

and

B

is

View Solution

Q5

In the network shown below, the equivalent resistance between P and Q is

\frac{4}{3} Ω

Hence the value of r is

View Solution