Show that in one time constant 63% of the equilibrium charge stored on the capacitor?
Answers
To show:
Charge accumulated in a Capacitor in one time constant is 63% of the total charge.
Calculation:
General expression for accumulation of charge in a capacitor in time "t" is ;
At one time constant ,
[Hence proved]
Explanation:
To show:
Charge accumulated in a Capacitor in one time constant is 63% of the total charge.
Calculation:
General expression for accumulation of charge in a capacitor in time "t" is ;
V = V_{0} \bigg \{1 - {e}^{( - \frac{t}{RC} )} \bigg \}V=V
0
{1−e
(−
RC
t
)
}
= > q =( CV_{0}) \bigg \{1 - {e}^{( - \frac{t}{RC} )} \bigg \}=>q=(CV
0
){1−e
(−
RC
t
)
}
= > q = q_{0} \bigg \{1 - {e}^{( - \frac{t}{RC} )} \bigg \}=>q=q
0
{1−e
(−
RC
t
)
}
At one time constant , t = \tau = RCt=τ=RC
= > q = q_{0} \bigg \{1 - {e}^{( - \frac{RC}{RC} )} \bigg \}=>q=q
0
{1−e
(−
RC
RC
)
}
= > q = q_{0} \bigg \{1 - {e}^{( - 1 )} \bigg \}=>q=q
0
{1−e
(−1)
}
= > q = q_{0} \bigg \{1 - \dfrac{1}{e} \bigg \}=>q=q
0
{1−
e
1
}
= > q = q_{0} \bigg \{1 - 0.37 \bigg \}=>q=q
0
{1−0.37}
= > q = q_{0} \bigg \{0.63 \bigg \}=>q=q
0
{0.63}
= > q =63\%( q_{0})=>q=63%(q
0
)
[Hence proved]