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In the circuit shown, the switch S is connected to position P for a long time so that the charge on the capacitor becomes $q_1$ μC. Then S is switched to position Q. After a long time, the charge on the capacitor is $q_2$ μC.

Q.1 The magnitude of $q_1$ is ___ .
Q.2 The magnitude of $q_2$ is ___ .

Solution

The figure below shows the situation when switch S is connected to position P.

No current flows through the capacitor in steady state.

So, $I=\frac {2-1}{1+2}=\frac {1}{3} A$

$V_C =2-2I=2-2\times \frac {1}{3}=\frac {4}{3} V$

$q_1 =CV_C =1\times \frac {4}{3} = \frac {4}{3} \mu C$

Actually, the answer to Q.1 is $q_1 =\frac {4}{3}=1.33$ as the unit is outside.

The figure below shows the situation when switch S is connected to position Q.

Again, no current flows through the capacitor in steady state.

$I=\frac {2}{1+2}=\frac {2}{3} A$

$V_C = \frac {2}{3} \times 1 = \frac {2}{3} V$

$q_2 = 1 \times \frac {2}{3} = \frac {2}{3} \mu C$

Actually, the answer to Q.2 is $q_2 =\frac {2}{3}=0.67$ as the unit is outside.

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$