Consider a capacitor-charging circuit. Let Q_1 be the charge give… - Vidyadip

MCQ

Consider a capacitor$-$charging circuit. Let $Q_1$ be the charge given to the capacitor in a time interval of $10 \ ms$ and $Q_2$ be the charge given in the next time interval of $10 \ ms$ . Let $10 \ mC$ charge be deposited in time interval $t _1$ and another $10 \ mC$ charge be deposited in the next time interval $t _2$.

A
$Q _1 > Q _2, t _1 > t _2$
✓
$Q _1> Q _2, t _1< t _2$
C
$Q _1< Q _2, t _1> t _2$
D
$Q _1< Q _2, t _1< t _2$

✓

Answer

Correct option: B.

$Q _1> Q _2, t _1< t _2$

Condition for charging capacitor-
$\text{Q}=\text{Q}_0\Big(1-\text{e}^\frac{-\text{t}}{\text{Rc}}\Big)$
$\text{Q}=\text{Q}_0\Big(1-\text{e}^\frac{-10\text{m}}{\text{Rc}}\Big)\ ...(1)$
$\text{Q}_1=\text{Q}_2=\text{Q}_0\Big(1-\text{e}^\frac{{-(10\text{m}+10\text{m})}}{\text{Rc}}\Big)$
$\text{Q}_1=\text{Q}_2=\text{Q}_0\Big(1-\text{e}^\frac{{-20\text{m}}}{\text{Rc}}\Big)\ ...(2)$
From eq. $(1)$ and $(2)$ we get-
$\text{Q}_1 > \text{Q}_2$
Given
$\text{Q}=\text{Q}_0\Big(1-\text{e}^\frac{-\text{t}}{\text{Rc}}\Big)$
$10\text{mc}=\text{Q}_0\Big(1-\text{e}^{\frac{-\text{t}}{\text{Rc}}}\Big)\ ...(3)$
$10\mu\text{c}+10\mu\text{c}=\text{Q}_0\Big(1-\text{e}^{\frac{-(\text{t}_1+\text{t}_2)}{\text{Rc}}}\Big)\rightarrow$
$20\mu\text{c}=\text{Q}_0\Big(1-\text{e}^{\frac{-(\text{t}_1+\text{t}_2)}{\text{Rc}}}\Big)\ ...(4)$
From eq. $(3)$ and $(4)$ we get,
$\text{t}_2 > \text{t}_1$

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