Question

A voltmeter of resistance R_1 and an ammeter of resistance R_2 are connected in series across a battery oif negligible internal resistance. When as resistance R is connected in parallel to voltmeter reading of ammeter increases three times white that of voltmeter reduces to one third. Find R_1 and R_2 in terms of R.

Answer 1

Answer:

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Answer 2

R1 and R2 in terms of R is :

• In first case, when R is not connected in parallel with voltmeter

• Applying Kirchoff's loop law

E - iR1 - iR2 = 0

E = i(R1 + R2)

R(total) = R(initial) = R1 + R2

i = E / R1+ R

• Second case : when R is connected in parallel with voltmeter

• Given : final current in circuit, i' = 3i and final voltage, v' = v/3

Potential across R1 and R will be same

But current through R1 = i/3

and current through R = 3i - i/3 = 8i/3

• Now potential across both resistance is same therefore,

i(R1)×R1 = i(R)×R

(i/3)×R1 = (8i/3)×R

R1 = 3R

• In main circuit, potential remains constant i.e given by v = iR

But i increases by 3 times, therefore R decreases by 1/3 times to maintain potential.

• Rf = (1/3)×Ri

Ri = R1 + R2 and Rf = R2 + (R1×R / R1+R)

• So,

R1 + R2 = R2 + (R1×R / R1+R)

Putting R1 = 3R

We get,

• R2 = 8R/3