wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find potential difference across the resistance 300 Ω given in circuit A and B.


A
60 V and 50 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
70 V and 70 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
30 V and 30 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
40 V and 30 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 60 V and 50 V
In (A):

Both the resistors are in series thus, potential difference across them directly proportional to their resistances.

V(300 Ω)=IR=100200+300×300=60 V

In (B):

300 Ω and voltmeter are in parallel, their effctive resistance is in series with 200 Ω

Total resistance RT is,

RT=200+300×600300+600=400 Ω

Total current through the battery is,

I=100400=0.25 A

As we know, across the parallel branches current gets divided in the inverse ratio of resistance of the branch.

I1:I2=1300:1600=2:1

I1=23×0.25=0.16 A

Hence potential difference across the 300 Ω is,

V=0.16×300=50 V

Alternate Solution:

300 Ω and 600 Ω are in parallel, thus potential difference across them will be the same.

V(300 Ω)=IR=100200+300×600300+600×[300×600300+600]=50 V

Hence, option (a) is the correct answer.
Why this question ?

We see that by the connected voltmeter the voltage which was to be measured has changed. Such voltmeters are not good. If its resistance had been very large than 300 Ω then it would have not affected the voltage by much amount.



flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon