Question

A triode valve has an anode resistance of 20,000 and an amplification factor of 20. The mutual conductance is:

• A. \( 10^{-2} \, \text{mho} \)
• B. \( 10^{-3} \, \text{mho} \)
• C. \( 10^3 \, \text{mho} \)
• D. \( 10^2 \, \text{mho} \)

Hint:

Mutual conductance \( g_m \) is inversely proportional to the anode resistance and directly related to the amplification factor.

Answer 1

The Correct Option is B

To determine the mutual conductance (\( g_m \)) of a triode valve, we need to use the relationship between the amplification factor (\( \mu \)), anode resistance (\( r_a \)), and mutual conductance. The formula is given by:

\( g_m = \frac{\mu}{r_a} \)

Here, we have:

• The amplification factor (\( \mu \)) is 20.
• The anode resistance (\( r_a \)) is 20,000 ohms.

Substituting these values into the formula gives:

\( g_m = \frac{20}{20000} \)

Simplifying the fraction:

\( g_m = \frac{1}{1000} = 10^{-3} \, \text{mho} \)

Thus, the mutual conductance is \( 10^{-3} \, \text{mho} \).

Answer 2

The mutual conductance \( g_m \) is given by the formula: \[ g_m = \frac{\mu}{R_a} \] Where:
- \( \mu \) is the amplification factor,
- \( R_a \) is the anode resistance. Substituting the values: \[ g_m = \frac{20}{20,000} = 10^{-3} \, \text{mho} \]

Answer 3

The mutual conductance of a triode valve equals the amplification factor divided by the anode resistance. Checking each option against the value of \( \frac{\mu}{r_a} \) tells us directly which one is correct.

1. \( 10^{-2} \, \text{mho} \): This equals \( 0.01 \). Since \( \frac{20}{20000} = 0.001 \), this option is ten times too large.
2. \( 10^{-3} \, \text{mho} \): This equals \( 0.001 \). Dividing the amplification factor by the anode resistance gives \( \frac{20}{20000} = 0.001 \), which matches this option exactly.
3. \( 10^{3} \, \text{mho} \): This equals \( 1000 \), which is far too large. Since the anode resistance (\( 20000 \, \Omega \)) is much bigger than the amplification factor (\( 20 \)), their ratio must be a small fraction well below \( 1 \), not a number in the thousands.
4. \( 10^{2} \, \text{mho} \): This equals \( 100 \), again far too large for the same reason, a small number divided by a much larger one cannot give a result above \( 1 \).

Since the amplification factor is a thousand times smaller than the anode resistance, their ratio comes out to a small fraction, and direct division confirms it is exactly \( 0.001 \, \text{mho} \).

Therefore, the correct answer is \( 10^{-3} \, \text{mho} \).