Current Electricity MSQ

Power maximum when r = R.
So, power consumed by it will decrease. for R > r.

The correct answers are: power consumed by it will increase, power consumed by it may increase or may decrease, power consumed will first increase then decrease.

The area of cross-section of conductor at point  A  is less than that at point B. So current density at  A  is higher. Hence, the electric field at  A  is more than at  B  and the thermal power generated at  A  is more than at  B  in an element of small same width, since resistance at A  is greater.

The correct answers are: The electric field at A  is more than at  B., The thermal power generated at  A  is more than at  B  in an element of small width.

(A)  Potential difference across each cell = V_P – V_Q

(B)  If i is clockwise then  E₂  is source and for anti-clockwise current  E₁  is source.

(C)  Potential difference = E – ir (when battery supplies energy) = E + ir  (when battery consumes energy).

The correct answers are: The potential difference across both the cell will be equal, One of the cell, will supply energy to the other cell., The potential difference across one of the cells will be greater than its emf.,

Power supplied by 20V cell = (–1)(20) = – 20W
as the cell is not supplying the power, it is eating the power (getting charged)

The correct answers are: current through 25V cell is 12.5A, power supplied by 20V cell is –20W

Let a  be the radius of left end side cross-section, then radius of cross-section at distance x  from left end is a + bx  where  b  is a constant.

 as i and σ are same for all cross-section

Rate of heat generation per unit length, 

The correct answers are:

there is zero potential difference across 4Ω and 6Ω resistance.

power by battery P_b = 

The correct answers are: power supplied by the battery is 200 watt, potential difference across 4Ω resistance is equal to the potential difference across 6Ω resistance

Total charge =  = Area under the curve = 10C

Maximum Power = I²R  when I  is maximum current = 100 × 1 = 100W

The correct answers are: Total charge flown through the resistor is 10C., Average current through the resistor is 5A., Maximum power during the flow of current is 100W.

As the length is doubled, the cross section are of the wire becomes half, thus the resistance of the wire  becomes four times the previous value. Hence after the wire is elongated the current becomes one fourth. Electric field is potential difference per unit length and hence becomes half the initial value. The power delivered to resistance is  and hence becomes one fourth.
The correct answers are: the electric field in the wire will become half., the thermal power produced by the resistor will become one fourth.

Resistance absorbs energy at the rate of 2W.

Potential difference across AB   V_AB·I = 50W

V_AB = 50V

Drop across resistor is 2V, therefore EMF of E is 48V.

As AB is absorbing energy at the rate of 50W, 48W  is being absorbed by  E. Thus E  is on charging mode i.e. current is entering from +ve terminal of  E.

The correct answers are: Emf of the device is 48V., Rate of conversion from electrical to chemical energy is 48W  in device  E.

Let the currents be as shown in the figure
KVL along ABCDA
⇒  –10i – 2 + (2 – i)1 = 0
∴  i = 0
Potential difference across S = (2 – i)1 = 2 × 1 = 2V

The correct answers are: The reading of the galvnometer is zero, The potential difference across the resistance S  is 2V