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**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 1Î© and to a resistance of 3Î© . 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 = 2

Ans:

= 12 Ã— 3 = 36 W

Power dissipated in resistance = i

= (3) (2)

**Introductory Exercise 20.5**

**Ques 1: Find the equivalent emf and internal resistance of the arrangement shown in Fig.Ans: **

r = 0.5 W

Ans:

P = power across R = i

For power to be maximum,

By putting we get, R = r

Further, by putting R = r in Eq. (i)

we get,

Ans:

Here, E = net emf = 2 + 2 = 4 V

and r= net internal resistance

= 1 + 1 = 2Î©

Ans:

= series resistance connected with galvanometer

Ans:

Ans:

Now, nV = i

(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:

or emf of lower battery

Solving this equation we get,

l = 320 cm

= 14r Now the circuit is as under,

Applying loop law in upper loop we have,

E - 14r(i

Applying loop law in lower law loop we have,

-E/2 - i

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 â†’ 4

Red â†’ 2

Orange â†’ 10

Gold â†’ 5

R = (4.2 Ã— 10

4 â†’ Yellow

10

5% â†’ Gold

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