Answer:
Battery voltage will be equal to 9.65 volt
Explanation:
We have given capacitance [tex]C=20\mu F=20\times 10^{-6}F[/tex]
Resistance [tex]R=6kohm=6000ohm[/tex]
Time constant of RC circuit is
[tex]\tau =RC=20\times 10^{-6}\times 6000=0.12sec[/tex]
Time is given t = 0.15 sec
Current i = 0.46 mA
Current in RC circuit is given by
[tex]i=\frac{V}{R}e^{\frac{-t}{\tau }}[/tex]
[tex]0.00046=\frac{V}{6000}e^{\frac{-0.15}{0.12 }}[/tex]
[tex]0.00046=\frac{V}{6000}\times 0.286[/tex]
V = 9.65 volt
So battery emf will be equal to 9.65 volt
The battery's emf will be "9.63 V".
Given values are:
Capacitor,
= [tex]20\times 10^{-6} \ F[/tex]
Resistor,
= [tex]6\times 10^3 \ \Omega[/tex]
At t = 0.15,
Current,
= [tex]0.46\times 10^{-3} \ A[/tex]
Since,
→ [tex]I = I_0.e^{-t/RC}[/tex]
→ [tex]I = \frac{E}{R}. e^{-t/RC}[/tex]
or,
→ [tex]E = \frac{IR}{e^{-t/Rc }}[/tex]
[tex]= \frac{(0.46\times 10^{-3}A)(6\times 10^3 \Omega)}{e^{-[\frac{0.15}{(6\times 10^3)(20\times 10^{-6})} ]}}[/tex]
[tex]= 9.63 \ V[/tex]
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