Question

a capacitor of capacitance C 1 is equal to 1 microfarad with stand a maximum voltage v1 is equal to 6 kv while another capacitance c 2 is equal to 2 microfarad with stand a maximum voltage v2 equal to 4 kv if they are connected in series what maximum voltage the system will with stand ??​

Answer 1

Answer:

Amount of charge, that the capacitor of capacitance $\sf{C_{1}}$ can withstand:

$\boxed{\sf{q_{1} = C_{1}V_{1}}}$

And:

Similarly the charge, that the capacitor of capacitance $\sf{C_{2}}$ can withstand:

$\boxed{\sf{q_{1} = C_{2}V_{2}}}$

But:

In series combination:

Charge on both the capacitors will be same, so, $\sf{q_{max}}$, that the combination can withstand $\sf{C_{1}V_{1}}$, as $\sf{C_{1}V_{2}}$ < $\sf{C_{2}V_{2}}$ , from the numerical data, given.

Now,

Net capacitance of the system:

$\boxed{\sf{C_{0} = \frac{C_{1}+ C_{2}}{C_{1} \times C_{2}}}}$

Hence:

$\sf{V_{max} = \frac{q_{max}}{C_{0}}}$

$\sf{V_{max} = \frac{C_{1}V_{1}}{ \frac{C_{1}C_{2}}{C_{1} + C_{2}} }}$

$\sf{V_{max} = V_{1}(1 + \frac{C_{1}}{C_{2}} )}$

Or:

$\sf{V_{max} = \frac{q_{1}}{C_{1}} + \frac{q_{1}}{C_{2}}}$

$\sf{V_{max} = \frac{6 \times {10}^{ - 3} }{1 \times {10}^{ - 6} } + \frac{6 \times {10}^{ - 3} }{2 \times {10}^{ - 6} }}$

$\sf{V_{max} = 6 \times {10}^{3} + 3 \times {10}^{3}}$

$\sf{V_{max} = {10}^{3} (6 + 3)}$

$\sf{V_{max} = 9 \times {10}^{3}\:V}$

$\sf{V_{max} = 9 \: kV}$

Thus,

If they are connected in series, the maximum voltage the system would with stand: 9 kV