Introductory Exercise 20.4
Ques 1: In the circuit shown in figure, a 12 V power supply with unknown internal resistance r is connected to a battery with unknown emf. E and internal resistance 1Ω and to a resistance of 3Ω . carrying a current of 2 A. The current through the rechargeable battery is 1 A in the direction shown. Find the unknown current i, internal resistance r and the emf E.
Ans: Applying loop law equation in upper loop we have,
E + 12  ir  1 = 0 ...(i)
Applying loop law equation in lower loop we have where i = 1 + 2 = 3A
E + 6  1 = 0 ...(ii)
Solving these two equations we get, E = 5Vand r = 2Ω
Ques 2: In the above example, find the power delivered by the 12 V power supply and the power dissipated in 3 W resistor.
Ans: Power delivered by a battery = Ei
= 12 × 3 = 36 W
Power dissipated in resistance = i^{2}R
= (3) (2)^{2} = 12W
Introductory Exercise 20.5
Ques 1: Find the equivalent emf and internal resistance of the arrangement shown in Fig.
Ans:
r = 0.5 W
Ques 2: If a battery of emf E and internal resistance r is connected across a load of resistance R. Show that the rate at which energy is dissipated in R is maximum when R = r and this maximum power is P = E^{2}/4r.
Ans:
P = power across R = i^{2}R
For power to be maximum,
By putting we get, R = r
Further, by putting R = r in Eq. (i)
we get,
Ques 3: Two identical batteries each of emf E = 2 volt and internal resistance r =1 ohm are available to produce heat in an external resistance by passing a current through it. What is the maximum power that can be developed across an external resistance R using these batteries.
Ans: As derived in the above question,
Here, E = net emf = 2 + 2 = 4 V
and r= net internal resistance
= 1 + 1 = 2Ω
Ques 4: The full scale deflection current of a galvanometer of resistance 1Ω is 5 mA. How will you convert it into a voltmeter of range 5 V ?
Ans: V = i_{g}(G + R)
= series resistance connected with galvanometer
Ques 5: A micrometer has a resistance of 100Ω and full scale deflection current of 50μA. How can it be made to work as an ammeter of range 5 mA ?
Ans:
Ques 6: A voltmeter has a resistance G and range V. Calculate the resistance to be used in series with it to extend its range to nV.
Ans: V = i_{g}G
Now, nV = i_{g}(G + R) = V/G (G + R) R = (n  1)G
Ques 7: The potentiometer wire AB is 600 cm long.
(a) At what distance from A should the jockey J touch the wire to get zero deflection in the galvanometer.
(b) If the jockey touches the wire at a distance 560 cm from A, what will be the current through the galvanometer.
Ans: (a)
or emf of lower battery
Solving this equation we get,
l = 320 cm
(b) Resistance of 560 cm =
= 14r Now the circuit is as under,
Applying loop law in upper loop we have,
E  14r(i_{1}  i_{2})  i_{1}r  i_{1}r = 0 ...(i)
Applying loop law in lower law loop we have,
E/2  i_{2}r + (i_{:} i_{2})(14r) = 0 ...(ii)
Solving these two equations we get
Introductory Exercise 20.6
Ques 1: For the given carbon resistor, let the first strip be yellow, second strip be red, third strip be orange and fourth be gold. What is its resistance?
Ans: Yellow → 4
Red → 2
Orange → 10^{3 }
Gold → 5
R = (4.2 × 10^{3} + 5%) Ω
Ques 2: The resistance of the given carbon resistor is (2.4 ×10^{6}Ω ± 5%)Ω . What is the sequence of colours on the strips provided on resistor?
Ans: 2 → Red
4 → Yellow
10^{6} → Blue
5% → Gold
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