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Physics Homework Sample Questions & Answers
An e.m.f. 10 volts is applied to a having a resistance of 10 ohms and an inductance henry. Find the time required by the current to 63.2% of its final value .What is the time constant circuits? Question 1.
I = I0 (1 - e-Rt/L)
Solution.
Given I/ I0 = 63.2 100
63.2 100
đ?‘…
;
đ??ż
=
10 0.5
= 20
= 1 - e-20t
e-20t = 1 - 0.632 = 0.368 e20t =
1 0.368
= 2.717 Copyright Š 2010-2014 Tutorhelpdesk.com
circuit of 0.5 attain of the
20t t= =
= loge2.717 1
20
x 2.3026 x log102.717
2.3.26 X 0.4341 20
= 0.05 sec
The time constant of the circuit is đ??ż đ?‘…
=
0.5 10
=
1 20
sec.
Question 2. An inductance of 500 mH and a resistance of
5 ohms are connected in serious with an e.m.f of 10 volts .Find the final current. If now the cell is removed and the terminals are connected together, find the current after (i) 0.05 sec. and (ii) 0.2 sec. Solution. Final current
I0 =
đ??¸ đ?‘…
=
10 5
= 2A
During discharge, I = I0 e-Rt/L Now
đ?‘… đ??ż
=
5 500 đ?‘‹ 10−3
= 10
(i) When t = 0.05 sec, I = 2e-10 *0.05 = 1.213A (ii) When t = 0.2 sec, I = 2e-10 *0.2
= 0.271A
Question 3. A capacitor is charged by DC supply through
a resistance of 2 megaohms. If it takes 0.5 seconds for the charge to reach three quarters of its final value, what is the capacitance of the capacitor?
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Solution. Here, R = 2 x 10
Q = Qo (1- e
–t/CR
)
Or or
đ?‘’
�, t = 0.5 s, Q/Qo =
or Q/Qo = ( 1- e 3 4
0.5 đ?‘?∗(2∗10 6 )
6
–t/CR
3 4
)
=
(1-đ?‘’
−0.5 đ?‘? x ( 2 x 10 6 )
)
or 0.5 /c x (2 x 106) = log đ?‘’ 4
= 4
or C = 0.5 / ( 2.3026 log10 4) (2 x 106) = 0.18 x 10-6 F = o.18 đ?œ‡F
Question 4. A capacitor of capacitance 0.1 đ?œ‡F is first
charged and then discharged through a resistance of 10 megaohm .Find the time, the potential will take to fall to half its original value. Solution. Q = Qo e đ?‘„đ?‘œ
đ?‘Ą
Q
đ??śđ?‘…
Or ln ( ) =
–t/CR
or
đ?‘„ đ?‘„đ?‘œ
= e
–t/CR
đ?‘„đ?‘œ
∴ t = CR ln ( ) đ?‘„
Q = CV and Qo = CV0
∴
đ?‘‰đ?‘œ
∴ t = CR ln ( ) đ?‘‰
Here, C = 10-7 F; R = 107 � ,
đ?‘‰đ?‘œ đ?‘‰
= 2.
∴ đ?‘Ą = 10-7 x 107 x ln 2 = 0.6931 s.
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đ?‘„đ?‘œ Q
=
đ?‘‰đ?‘œ đ?‘‰
Question 5. A resistance R and a 2đ?œ‡F capacitor in series
are connected to a 200 volt direct supply. Across the capacitor is a neon lamp that strikes at 120 volts. Calculate the value of R to make the lamp strikes 5 seconds after switch has been closed.
Solution. The resistance R must be such that the p.d
across the capacitor should rise to 120 volt in 5 seconds after the switch is closed .The lamp would then strike. The equation of charging is
Q = Qo (1 -
–t/CR ) e
Or CV = CV0 (1 Or
V = V0 (1 -
–t/CR e –t/CR e
)
)
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Here, V =120 volt, V0 = 200 volt, t = 5 sec and C = 2 x 10-6 farad, ∴ 120 = 200 (1Or
–5/2x10-6 e
–5/2x10-6 ) e
= 5/2
5/2 x 10-6R = loge5 – log e 2 =1.6094 - 0.6931 =0.9163 R = 5/2 x 10-6 x 0.9163 = 2.73 x 106 ohm = 2.73 megaohm
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