CURRENT O12 - CHATUP707

Question 12 Solution

Step 1: Define the Relationship

Let the initial Voltmeter reading be V₁ and Ammeter reading be A₁.

In the second case (with parallel resistance), the readings change by a factor of η:

A 2 = η A 1 V 2 = V 1 η

Using Ohm’s Law (V = IR) for the voltmeter section, we find the new equivalent resistance R’_v:

R’ v = V 2 A 2 = V 1 η η A 1 = V 1 η 2 A 1 = R v η 2

Step 2: Circuit Equations

Let R_x be the total series resistance (internal resistance + ammeter resistance).

The total current ratio is inversely proportional to the total circuit resistance:

A 2 A 1 = η = R x + R v R x + R’ v

Substitute R’_v = R_v / η²:

R x + R v = η (R x + R v η 2)

Step 3: Solve for R_x

Expand and rearrange the equation:

R x + R v = η R x + R v η

R v - R v η = η R x - R x

R v ( 1 - 1 η) = R x (η - 1)

R v (η - 1 η) = R x (η - 1)

Canceling (η – 1) from both sides gives us a crucial relationship:

R x = R v η

Step 4: Calculate Initial Reading

The initial reading V₁ is the potential drop across R_v in the original series circuit.

V 1 = Ε (R v R x + R v)

Substitute R_x = R_v / η:

V 1 = Ε (R v R v η + R v)

Divide numerator and denominator by R_v:

V 1 = Ε (1 1 η + 1)

Multiply numerator and denominator by η:

V 1 = η Ε 1 + η

Answer: (b)

ηΕ η + 1