Electricity Worksheet-1
A. 1/25 B. 1/5 C. 5 D. 25
A. I^{2}R B. IR^{2} C. VI D. V^{2}/R
A. 100 W B. 75 W C. 50 W D. 25 W
A. 1:2 B. 2:1 C. 1:4 D. 4:1
A. volt B. joule C. coulomb D. ohm.
A. +1.6 × 10^{–29} C B. +1.6 × 10^{–9} C
C. –1.6 × 10^{–9} C D. +10^{10} C
A. one-half B. double
C. one-fourth D. four times.
A. 1.6 × 10^{–19} C B. 3.2 × 10^{–19} C
C. 1 C D. 3 × 10^{9} C
A. 1.6 × 10^{–19} C B. 3.2 × 10^{–19} C
C. 6.4 × 10^{–19} C D. 0.8 × 10^{–19} C
A. 1.6 × 10^{–26} A B. 1.6 × 10^{12} A
C. 1.6 × 10^{–12} A D.
Answer Key:
Explanation: Resistance of each one of the five parts = (R/5)
Resistance of five parts connected in parallel is given by
or
or
Explanation: Electrical power, P = VI = (IR)R = I^{2}R
IR^{2} does not represent electrical power in a circuit.
Explanation: Resistance of the electric bulbs,
or
Power consumed by the bulb when it is operated at 110 V
Explanation: Since both the wires are made of the same material and have equal lengths and equal diameters, so they will have the same resistance.
Assume it to be R.
When connected in series, their equivalent resistance is given by
R_{s} = R + R = 2R
When connected in parallel, their equivalent resistance is given by
or
Further, electrical power is given by
Power (or heat produced) in series,
Power (or heat produced) in parallel,
or R_{s} : R_{p} :: 1 : 4