
The amplification factor of a triode is 50. If the grid potential is decreased by V. What increase in plate potential will keep the plate current unchanged?
(A) 5 V
(B) 10 V
(C) 0.2 V
(D) 50 V
Answer
562.8k+ views
Hint:
The concept of the given question is based on the amplification of a triode. The formula for calculating the amplification factor, $\mu = \dfrac{{\Delta {V_P}}}{{\Delta {V_g}}}$, is used to solve this problem. As the values of the amplification factor and change in the grid voltage are given, substituting these values in the formula, the plate voltage, that is the required value can be obtained.
$\Rightarrow \mu = \dfrac{{\Delta {V_P}}}{{\Delta {V_g}}}$
$\Delta {V_P}$ is the change in the plate to cathode voltage and$\Delta {V_g}$ is the change in the grid to cathode voltage.
and $\mu $ is the amplification factor.
Complete step by step answer:
The amplification factor is given by a formula as follows:
$\Rightarrow \mu = \dfrac{{\Delta {V_P}}}{{\Delta {V_g}}}$
where ${V_P}$ is the plate to cathode voltage and ${V_g}$ is a grid to cathode voltage.
We are given in the question the value of the amplification factor of a triode to be 50 and the value of the grid potential to be V.
So, let us substitute these values in the formula, which is used to find the value of the amplification factor, to determine the change in the value of the plate voltage.
So, we have,
$\Rightarrow \Delta {V_P} = - \mu \times \Delta {V_g}$
Now substituting the value gives us,
$\Rightarrow \Delta {V_P} = - (50) \times ( - 0.20)$
On calculation we get,
$\Rightarrow \Delta {V_P} = + 10\,V$
Hence this is required to change the value of the plate voltage.
$\therefore $ In order to keep the plate current unchanged, the increase in the value of the plate voltage should be 10 V, thus, option (B) is correct.
Note:
The negative sign is used for grid potential, as it is given in the question that, this voltage value decreased by some amount. So, to find the increase in the value of the plate voltage, the amplification factor is also considered to be negative.
The concept of the given question is based on the amplification of a triode. The formula for calculating the amplification factor, $\mu = \dfrac{{\Delta {V_P}}}{{\Delta {V_g}}}$, is used to solve this problem. As the values of the amplification factor and change in the grid voltage are given, substituting these values in the formula, the plate voltage, that is the required value can be obtained.
$\Rightarrow \mu = \dfrac{{\Delta {V_P}}}{{\Delta {V_g}}}$
$\Delta {V_P}$ is the change in the plate to cathode voltage and$\Delta {V_g}$ is the change in the grid to cathode voltage.
and $\mu $ is the amplification factor.
Complete step by step answer:
The amplification factor is given by a formula as follows:
$\Rightarrow \mu = \dfrac{{\Delta {V_P}}}{{\Delta {V_g}}}$
where ${V_P}$ is the plate to cathode voltage and ${V_g}$ is a grid to cathode voltage.
We are given in the question the value of the amplification factor of a triode to be 50 and the value of the grid potential to be V.
So, let us substitute these values in the formula, which is used to find the value of the amplification factor, to determine the change in the value of the plate voltage.
So, we have,
$\Rightarrow \Delta {V_P} = - \mu \times \Delta {V_g}$
Now substituting the value gives us,
$\Rightarrow \Delta {V_P} = - (50) \times ( - 0.20)$
On calculation we get,
$\Rightarrow \Delta {V_P} = + 10\,V$
Hence this is required to change the value of the plate voltage.
$\therefore $ In order to keep the plate current unchanged, the increase in the value of the plate voltage should be 10 V, thus, option (B) is correct.
Note:
The negative sign is used for grid potential, as it is given in the question that, this voltage value decreased by some amount. So, to find the increase in the value of the plate voltage, the amplification factor is also considered to be negative.
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