A. 400 Q
B. 600 Q
C. 800 Q
D. 1000 Q
Advertisement
Related Mcqs:
- A 40 W bulb is connected in series with a room heater. If now 40 W bulb is replaced by 100 W bulb, the heater output will__________
A. decrease
B. increase
C. remain same
D. heater will burn out - The hot resistance of the bulb’s filament is higher than its cold resistance because the temperature co-efficient of the filament is___________________?
A. zero
B. negative
C. positive
D. about 2 ohms per degree - The resistance of a parallel circuit consisting of two branches is 12 ohms. If the resistance of one branch is 18 ohms, what is the resistance of the other ?
A. 18 Q
B. 36 Q
C. 48 Q
D. 64 Q - Three 60 W bulbs are in parallel across the 60 V power line. If one bulb burns open_________________?
A. there will be heavy current in the main line
B. rest of the two bulbs will not light
C. all three bulbs will light
D. the other two bulbs will light - In a circuit a 33 Q resistor carries a current of 2 A. The voltage across the resistor is
A. 33 V
B. 66 v
C. 80 V
D. 132 V - Two resistances Rl and Ri are connected in series across the voltage source where Rl>Ri. The largest drop will be across______________?
A. Rl
B. Ri
C. either Rl or Ri
D. none of them - When a voltage of one volt is applied, a circuit allows one micro ampere current to flow through it. The conductance of the circuit is___________?
A. 1 n-mho
B. 106 mho
C. 1 milli-mho
D. none of the above - Internal resistance of ideal voltage source is ?
A. zero
B. infinite
C. finite
D. 100 ohms - The resistance of a conductor of diameter d and length l is R Ω. If the diameter of the conductor is halved and its length is doubled, the resistance will be
A. R Ω
B. 2R Ω
C. 4R Ω
D. 8R Ω - Copper wire of certain length and resistance is drawn out to three times its length without change in volume, the new resistance of wire becomes__________?
A. 1/9 times
B. 3 times
C. 9 times
D. unchanged
Advertisement
2 Comments
As we know
V=240
I=300mA
then what will be the resistance
R=?
According to formula
R=V/I
R=240/300mA
R=240×1000/300
R=2400/3
R=800