Two bulbs rated 200 W, 220V and 100 W, 220V, are connected in series, combination is connected to 220 V supply. Power consumed by the circuit is

• 8O W

• 67 W

• 76 W

• 65 W

In a potentiometer experiment, the balancing with a cell is at 240 cm. On shunting the cell with a resistance of 2 Ω the balancing length becomes 120 cm. The internal resistance of the cell is

• 0.5 Ω 

• 1 Ω 

• 2 Ω 

• 4 Ω 

The total current supplied to the circuit by the battery is

• 1 A

• 2 A

• 4 A

• 5 A

The current in a wire varies with time as I = 5t² + 2t, where I is in ampere and t is in second. The quantity of charge which has to be passed through a cross-section of wire during the times t = 4 s to t = 6 s is

• 220 C

• 340 C

• 273.34 C

• 280 C

Three resistances of 2 Ω, 3 Ω and 6 Ω are connected in parallel and the combination is connected in series with 2 V battery having internal resistance of 1 Ω . The current in the battery is

• 1 A

• 3 A

• 2 A

• 4 A

An electric cable of copper has just one wire of radius 9 mm with resistance 5 Ω . This single copper wire of cable is replaced by 4 different well insulated copper wires each of radius 3 mm. The total resistance of the cable now will be

• 22.5 Ω

• 45 Ω

• 9 Ω

• 11.25 Ω

387.

A rise of temperature 5°C is observed in a conductor by passing a current. If the current is doubled the rise in temperature will be

• 10°C

• 20°C

• 5°C

• 25°C

The current density varies with radial distance r as J =ar² (where, a is a constant) in a  cylindrical wire of radius R. The current passing through the wire between radial distance $\frac{\mathrm{R}}{4}$ and $\frac{\mathrm{R}}{2}$ is 

• $\frac{20{\mathrm{\pi aR}}^4}{331}$

• $\frac{15{\mathrm{\pi aR}}^4}{512}$

• $\frac{{\mathrm{\pi aR}}^4}{332}$

• $\frac{3{\mathrm{\pi aR}}^4}{864}$

A thermister is dipped in a water bath. As the temperature increases the current in the thermistor also increases. This is due to decrease in

• resistivity

• conductivity

• reactance

• capacitance

In an AC circuit, voltage and current are given by V = 50 sin (lO0 t) V and I = 50 sin $\left(100 \mathrm{t} + \frac{\mathrm{\pi }}{3}\right)$ A. The power dissipiated in the circuit is

• 125 W

• 625 W

• 1250 W

• 100 W