In a triode valve,the amplification factor is $20$ and the mutual conductance is $10^{-3} \text{ mho}$. The plate resistance is:
- A$2 \times 10^{3} \ \Omega$
- B$4 \times 10^{3} \ \Omega$
- C$2 \times 10^{4} \ \Omega$
- D$2 \times 10^{5} \ \Omega$
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Similar Questions
The linear portions of the characteristic curves of a triode valve give the following readings. The plate resistance is ....... $\Omega$.
$V_g$ (volt) $0$ $-2$ $-4$ $-6$ $I_p$ (mA) for $V_p = 150$ $V$ $15$ $12.5$ $10$ $7.5$ $I_p$ (mA) for $V_p = 120$ $V$ $10$ $7.5$ $5$ $2.5$
| $V_g$ (volt) | $0$ | $-2$ | $-4$ | $-6$ |
| $I_p$ (mA) for $V_p = 150$ $V$ | $15$ | $12.5$ | $10$ | $7.5$ |
| $I_p$ (mA) for $V_p = 120$ $V$ | $10$ | $7.5$ | $5$ | $2.5$ |
Difficult
View SolutionIn the space charge limited region,the plate current in a diode is $10\, mA$ for a plate voltage of $150\, V$. If the plate voltage is increased to $600\, V$,then the plate current will be.......$mA$.
Medium
View SolutionThe amplification factor of a triode is $20$. Its plate resistance is $10 \, k\Omega$. The mutual conductance is:
Easy
View SolutionCoating of strontium oxide on a tungsten cathode in a valve is good for thermionic emission because:
Easy
View SolutionThe relation between $I_p$ and $V_p$ for a triode is $I_p = (0.125V_p - 7.5) \text{ mA}$. Keeping the grid potential constant at $1 \text{ V}$,the value of $r_p$ will be . . . . . . $\text{k}\Omega$.
Medium
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