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Consider the circuit at t=0s, the kety K...

Consider the circuit at t=0s, the kety K is closed . At t=1 s, the current is l(t), Graph (1) is the variation of `log_(e) l(t)` with time 't'. To obtain graph (2) , one of the parameters that is V, R or C is charged keeping the other two constant , What should be the change, to obtain graph (2) from graph (1) ?

Text Solution

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`q_((t))=CV[1-e^(-t//eR)]`
Charging current `i_((t))= (d)/(gt)q(t) = (V)/(R ) e^((t)/(cR))`
`:. logl_((t)) = log.(V)/(R ) - ((1)/(CR )) .t `
This is the equation of straight line with intercept `log_(e ) .(V)/(R )` upon y-axis and negative slope, to obtain graph (2) 'c' should be increased. Because intercept has to remain constant and magnitude of slope has to decrease.
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