Question

Resistances of 90 and 61 are connected as shown in
the figure below. A 6 volt supply source is connected
across P and Q. The reading of ammeter (shown in
figure) is obtained Z milli-ampere. Find value of Z.
So
69
60
90
WW
992
w
mu
Р
69
692
Q
WWW
w
922
692
4.)
K
HUH
6V
Enter Your Answer​

Answer 1

Explanation:

Voltage drop across voltmeter is 3 V and emf is 4 V.

Hence, voltage across A

I

is 1 V.

Hence, resistance of ammeters is 100 Ω.

Let the potential of the point of intersection of the 100 Ω resistances and A

II

be x.

Using KCL:

100

x−4

+

100

x

+

100

x−1

=0

3x=5

⇒x=

3

5

Hence, current in A

II

=

100

x−1

=

300

2

=6.67 mA