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Alternating Current question

2011 · Shift 0 · Q58

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Alternating Current
2011 · Shift 0 · Q58

Alternating Current question

2011 · Shift 0 · Q58

JEE MainPhysicsAlternating CurrentMCQ+4 / −1

A resistor

'R'

and

2\mu F

capacitor in series is connected through a switch to

200V

direct supply. Across the capacitor is a neon bulb that lights up at

120V.

Calculate the value of

R

to make the bulb light up

5s

after the switch has been closed.

\left( {{{\log }_{10}}2.5 = 0.4} \right)

A
$1.7 \times {10^5}\,\Omega$
B
$2.7 \times {10^6}\,\Omega$
C
$3.3 \times {10^7}\,\Omega$
D
$1.3 \times {10^4}\,\Omega$

View written solutionFree

Correct answer: B

Charging equation of capacitor

For a capacitor charging through a resistor from a DC source $V_0$ , the capacitor voltage at time $t$ is

$V_C = V_0\left(1-e^{-t/RC}\right).$

Here,

$V_0 = 200\,\text{V}$
$C = 2\,\mu\text{F} = 2\times 10^{-6}\,\text{F}$
neon bulb lights when $V_C = 120\,\text{V}$
required time $t = 5\,\text{s}$

So,

$120 = 200\left(1-e^{-5/(RC)}\right).$

Solve for the exponential term

Divide by $200$ :

$\frac{120}{200} = 1-e^{-5/(RC)}$

$0.6 = 1-e^{-5/(RC)}$

$e^{-5/(RC)} = 0.4.$

Take logarithm

Using common logarithm:

$-\frac{5}{RC}\log_{10} e = \log_{10}(0.4).$

But it is easier to write

$\frac{5}{RC} = \ln\left(\frac{1}{0.4}\right)=\ln(2.5).$

Given:

$\log_{10}(2.5)=0.4.$

Hence,

$\ln(2.5)=2.303\log_{10}(2.5)=2.303\times 0.4\approx 0.9212.$

Thus,

$\frac{5}{RC}=0.9212$

$RC=\frac{5}{0.9212}\approx 5.43\,\text{s}.$

Calculate $R$

$R=\frac{RC}{C}=\frac{5.43}{2\times 10^{-6}}$

$R=2.715\times 10^6\,\Omega.$

So,

$R\approx 2.7\times 10^6\,\Omega.$

Check options

A: $1.7\times 10^5\,\Omega$ ✗
B: $2.7\times 10^6\,\Omega$ ✓
C: $3.3\times 10^7\,\Omega$ ✗
D: $1.3\times 10^4\,\Omega$ ✗

Therefore, the correct option is B.

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