First slide

combination of resistance

Question

Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 ohms is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohms is

Moderate

A

3
B

6
C

9
D

12

Solution

Let S be larger and R be smaller resistances connected in two gaps of meter bridge.
$∴ S = (\frac{100 - l}{l}) R = (\frac{100 - 20}{20}) R = 4 R$ ….(i)
When $15 Ω$ resistance is added to resistance R, then
$S = (\frac{100 - 40}{40}) (R + 15) = \frac{6}{4} (R + 15)$ ….(ii)
From equations (i) and (ii), $R = 9 Ω$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App