Question

Choose the correct alternative and rewrite the following:

(1) 1mA =_______ A

(a) 10³A

(b) 10⁻³ A

(c) 10⁶A

(d) 10⁻⁶ A

(2) To increase the effective resistance in a circuit the resistors are connected in _______

(a) series

(b) parallel

(c) both ways

(d) none of these

(3) 1 kilowatt hr = _______ Joules

(a) 4.6 × 10⁶ Joule

(b) 3.6 × 10⁶ Joule

(c) 30.6 × 10⁶ Joule

(d) 3.6 × 10⁵ Joule

(4) If a P. D. of 12 V is applied across a 3Ω resistor, then the current passing through it is _______

(a) 36 A

(b) 4 A

(c) 0.25 A

(d) 15 A

Answer 1

(1) 1mA = 10^-3 A

Description: Small currents are expressed in milli ampere where milli stands for 10^-3

i.e., 1mA = 10^-3 A.

(2) To increase the effective resistance in a circuit the resistors are connected in series.

Description: The effective resistance in a series combination is greater than the individual resistance and is given by:

R_S = R₁ + R₂ + R_3.

(3) 1 kilowatt hr = 3.6 × 10⁶ Joules

Description: The commercial unit of energy is kilowatt per hour (kWh) which means the amount of energy consumed in one hour.

1 kilowatt hour = 1000 Watt-hr

(4) If a P. D. of 12 V is applied across a 3Ω resistor, then the current passing through it is 4A

Description:

Given: Potential difference, V = 12V

Resistance, R = 3 Ω

From ohm’s law which gives relation between V and I as, V = IR

⇒ I = V/R

⇒ I = 12/3 (putting the given values)

⇒ I = 4A