The current flowing through a wire depends on time as I = 3t2 + 2t + 5. The charge flowing through the cross-section of the wire in time from t = 0 to t = 2 s is
22 C
20 C
18 C
5 C
Two cells each of same emf e but of internal resistances r1 and r2 are connected in series through an external resistance R. If the potential difference between the ends of the first cell is zero, what will be the value of R in terms of r1 and r2 ?
A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire
nR
n2R
2nR
2n2R
B.
n2R
The resistance of a wire is directly proportional to the length of the wire. Thus, R ∝ l, so the length increases n-fold also since the volume of the wire remains constant, so the area of the wire decreases n-fold. Hence the resistance increases to n2R as the area is inversely proportional to resistance.
A battery of emf E and internal resistance r is connected across a pure resistive device (such as an electric heater) of resistance R. Prove that the power output of the device will be maximum if R = r.
Five equal resistance, each of resistance R are connected as shown in figure below. A battery of V volt is connected between A and B. The current flowing in FC will be
Two cells with the same emf E and different internal resistances r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero, is
r1 + r2
r1 − r2
What current will flow through the 2 k Ω resistor in the circuit shown in the figure ?
3 mA
6 mA
12 mA
36 mA
The I- V characteristics of a metal wire at two different temperatures (T1 and T2) are given in the adjoining figure. Here, we can conclude that
T1 > T2
T1 < T2
T1 = T2
T1 = 2T2