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The plate current i(p) in a triode valve...

The plate current `i_(p)` in a triode valve is given `i_(p)=K(V_(p)+mu V_(g))^(3//2)` where `i_(p)` is in milliampere and `V_(p)` and `V_(g)` are in volt. If `r_(p)=10^(4) ohm` and `g_(m)=5xx10^(-3)mho`, then for `i_(p)=8mA` and `V_(p)=300` volt, what of K and grid cur off voltage

A

`-6V,(30)^(3//2)`

B

`-6V,(1//30)^(3//2)`

C

`+6V,(30)^(3//2)`

D

`+6V,(1//30)^(3//2)`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • The plate current in a triode is given by I_(p)=0.004(V_(p)+10V_(g))^(3//2)mA where I_(p), V_(p) and V_(g) are the values of plate current, plate voltage and grid voltage, respectively. What are the triode parameters mu r_(p) and g_(m) for the operating point at volt V_(p)=120 volt and V_(g) =-2 volt ?

    A
    `10,16.7 k Omeg, 0.6 m mho`
    B
    `15,16.7 k Omega, 0.06 m mho`
    C
    `20, 6 k Omega, 16.7 m mho`
    D
    None of these

  • The table values of E_(b) , E_(c) , I_(b) for a triode The value of r_(p) in k Omega and g_(m) in mA/V are given by

    A
    50 and 0.1
    B
    10 and 3.0
    C
    5 and 1.5
    D
    3.33 AND 1.5

  • For a triode r_(p)=10 kilo ohm and g_(m)=3 milli mho . If the load resistance is double of plate resistance, then the value of voltage gain will be

    A
    10
    B
    20
    C
    15
    D
    30

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