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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Current through Gases

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current through Gases

A triode valve operates with V_p = 225 V and V_g = -0.5 V . If the...

A triode valve operates with Vp=225 VV_p = 225 \text{ V} and Vg=−0.5 VV_g = -0.5 \text{ V}. If the plate voltage is raised to 250 V250 \text{ V} and the grid voltage is lowered to −2.5 V-2.5 \text{ V}, the plate current is found to remain unaltered. Evaluate the amplification factor.

Options

  1. A5050
  2. B1010
  3. C2525
  4. D12.512.5

Correct answer

D. 12.512.5

Step-by-step solution

The amplification factor μ\mu of a triode is given by the ratio of the change in plate voltage to the change in grid voltage, keeping the plate current constant.

μ=−(ΔVpΔVg)ΔIp=0\mu = - \left( \dfrac{\Delta V_p}{\Delta V_g} \right)_{\Delta I_p = 0}

Given Vp1=225 VV_{p1} = 225 \text{ V}, Vg1=−0.5 VV_{g1} = -0.5 \text{ V} and Vp2=250 VV_{p2} = 250 \text{ V}, Vg2=−2.5 VV_{g2} = -2.5 \text{ V}.

Change in plate voltage, ΔVp=Vp2−Vp1=250−225=25 V\Delta V_p = V_{p2} - V_{p1} = 250 - 225 = 25 \text{ V}

Change in grid voltage, ΔVg=Vg2−Vg1=−2.5−(−0.5)=−2.0 V\Delta V_g = V_{g2} - V_{g1} = -2.5 - (-0.5) = -2.0 \text{ V}

Substituting the values:

μ=−(25−2.0)=12.5\mu = - \left( \dfrac{25}{-2.0} \right) = 12.5
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