A battery of internal resistance 2Ω is connected to a variable resistor whose value can vary from 4Ω to 10Ω. The resistance is initially set at 4Ω. If the resistance is now increased then, which of the statement is not correct.
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.
In the figure a conductor of nonuniform crosssection is shown. A steady current I flows in it.
The area of crosssection 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.
Two cells of unequal emfs E_{1} and E_{2} and internal resistances r_{1} and r_{2} are joined as shown in figure. V_{P} and V_{Q} are the potential at P and Q respectively.
(A) Potential difference across each cell = V_{P} – V_{Q}
(B) If i is clockwise then E_{2} is source and for anticlockwise current E_{1} 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.,
In the figure shown: (All batteries are ideal)
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
A conductor of truncated conical (frustum) is connected to a battery of emf ∈ as shown in the figure.
If at a section distance x from left end, electric field intensity, potential and rate of generation of heat per unit length are E, V and H respectively, then which of the following graphs(s) is/are correct?
Let a be the radius of left end side crosssection, then radius of crosssection at distance x from left end is a + bx where b is a constant.
as i and σ are same for all crosssection
Rate of heat generation per unit length,
The correct answers are:
In the circuit shown in the figure
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
A variable current flows through a 1Ω resistor for 2 seconds. Time dependence of the current is shown in the graph.
Total charge = = Area under the curve = 10C
Maximum Power = I^{2}R 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.
Consider a resistor of uniform cross sectional area connected to a battery of zero internal resistance. If the length of the resistor is doubled by stretching it then :
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.
AB is part of a circuit as shown, that absorbs energy at a rate of 50W. E is an emf device that has no internal resistance.
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.
The galvanometer shown in the figure has resistance 10Ω. It is shunted by a series combination of a resistance S = 1Ω and an ideal cell of emf 2V. A current 2A passes as shown.
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
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