In the circuit shown in figure when S1 is closed and S2 is open, the ideal voltmeter shown a reading 18 V. When switch S2 is closed and S1 is open, the reading of the voltmeter is 24 V, when S1 and S2 both are closed, the voltmeter reading will be
A
14.4 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
20.6 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
24.2 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 14.4 V
Hello Amal, let r be the internal resistance of the cell and E is the e m f of the source
So V∗(r+R)=R∗E
Now in first case R=6 ohm and V=18V
Second case R=12 ohm and V=24V
So 18∗(r+6)=6∗E --(1)
And 24∗(r+12)=12∗E --(2)
Dividing (2) by (1) we get, 4/3∗(r+12)/(r+6)=2
===>2(r+12)=3∗(r+6)
Or r = 6 ohm
Now we can find E as 36 V
So when both switches closed then equivalent resistance will be 6∗12/(6+12)=4ohm
Hence the voltmeter reading will be got from ER=V(r+R)