CBSE
A Wheatstone bridge has the resistances 10 Ω, 10 Ω, 10 Ω and 30 Ω in its four arms. What resistance joined in parallel to the 30 Ω resistance will bring it to the balanced condition ?
2 Ω
5 Ω
10 Ω
15 Ω
A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be
2 Ω
4 Ω
8 Ω
16 Ω
The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 Ω, 30 Ω, 30 Ω and 90 Ω, respectively. The emf and internal resistance of the cell are 7 V and 5 Ω respectively. If the galvanometer resistance is 50 Ω, the current drawn from the cell will be
1.0 A
0.2 A
0.1 A
2.0 A
In the circuit shown assume the diode to be ideal. When Vi increases from 2V to 6 V, the change in the current is (in mA)
zero
20
80/3
40
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
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
A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is
once per revolution
twice per revolution
four times per revolution
six times per revolution
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
0.2 Ω
0.5 Ω
0.8 Ω
1.0 Ω
Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is
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