# Two resistors 400 ohm and 800 ohm are connected in series with a 6V battery. It isdesired to measure the current in the circuit. An ammeter of 10 ohm resistance is usedfor this purpose. What will be the reading in the ammeter? Similarly if a voltmeter of10,000 ohm resistance is used to measure the potential difference across 400 ohm, whatwill be the reading of the voltmeter ?

Nirmal Singh.
10 years ago

Regards,
Nirmal SIngh
Rajbhanu Shukla
24 Points
6 years ago
Dear Student
Total resistance of circuit is 400 +800 +10 ohm =1210 ohm (As resistors are connected in series with ammeter)
Current through circuit or each resistor or ammeter will be I = V/R = 0.0049 A = 4.9 x 10-3 A
Now when voltmeter is connected across 400 ohm resistance.
Then total resistance of voltmeter and 400 ohm resistor is

R = 384. 6 ohm
Current I through circuit = = 0.0051 A 6
384/ 6+400
Now let current passes through voltmeter and passes through 400 ohm resistor. I1 I2 V oltage across both
will be same.
So
400 I2 = 10000I1
I1 = 0.04 I2
But
I = I1 +I2
I2 +0.04 I2 = 0.0051
I2 = = 0.0049
0.0051
1.04
THerefore potential differecne across 400 ohm = 400 I2 = 400×0.0049 = 1.95 V
Therefore reading of ammeter = 1.95 V olt
Rishi Sharma
3 years ago
Dear Student,

Given,
R1 ​= 400Ω, R2 ​= 800Ω
E = 6V
1stCase
r = 10Ω
i = 6/(10 + 400 + 800) ​= 6/(1210) ​= 4.96mA
2ndCase
r = 10000Ω
Req​ = (10000×400)/10400​ + 800 = 30800​/26
i = 6×26/30800​
i 400Ω ​= (25/26​)×(6×26/30800)​
V 400Ω ​= i×400 = 150​×400/30800 = 150/77 ​= 1.95V
Hence option B is the correct answer.

Thanks and Regards