Question: Q. 21. A cell of emfE and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations :

(i) without any external resistance in the circuit

(ii) with resistance R1 only

(iii) with R1 and R2 in series combination

(iv) with R1 and R2 in parallel combination.

The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order. Identify the currents corresponding to the four cases mentioned above. A[Delhi I, II, III 2012]

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Solution:

Ans. Use the formulae and values come as follows :

(i) i=Er

(ii) i=Er+R1

(iii) i=Er+R1+R2

(iv) i=Er+R1R2R1+R2

(i) i=4.2 A

(ii) i=1.05 A

(iii) i=0.42 A

(iv) i=1.4 A AI Q. 22. In the circuit shown in the figure, find the current through each resistor.

A [Delhi I, II, III 2015]

Ans. Total emf in the circuit =8 V4 V=4 V1/2

Total resistance of the circuit =8Ω1/2

Hence current flowing in the circuit

=VR=48 A=0.5 A

Current flowing through the resistors :

Current through 0.5Ω,1.0Ω and 4.5Ω is 0.5 A

Current through 3.0Ω is 23I=13 A

Current through 6.0Ω is 13I=16 A [AI Q. 1. Derive the expression for the current density of a conductor in terms of the conductivity and applied electric field. Explain, with reason how the mobility of electrons in a conductor changes when the potential difference applied is doubled, keeping the temperature of the conductor constant. A [Delhi Comptt. I, II, III 2017]

Ans. Derivation of current density

Explanation with reason the change in mobility of electrons

Using Ohm’s law,

V=IR=IρlA

Potential difference (V), across the ends of a conductor of length ’ l ‘, where field ’ E ’ is applied is given by

V=El El=IρlA

But current density,

J=IA

El=Jρl=Jlσ [1ρ=σ]1/2 μ=vdE vd=eVτml

As potential is doubled, drift velocity also gets doubled, therefore, no change in mobility.

[CBSE Marking Scheme 2017]



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