Question

A resistor of 4-ohm resistance and an electric lamp of 20-ohm resistance are connected in series across a 6-volt battery. Draw a labelled circuit diagram by using the symbols for ohm and volt. Calculate
(a) total resistance of the circuit.
(b) current flowing through the circuit.
(c) potential difference across the resistor.
​(d) potential difference across the electric lamp.

Answer 1

Conclusion:

(a) total resistance of the circuit $= 24 \Omega$  

(b) current flowing through the circuit $= \textrm{ 0.25 amp}$

(c) potential difference across the resistor$= \textrm{1 Volt}$

​(d) potential difference across the electric lamp$= \textrm{5 Volt}$

Explanation:

Given that, resistance of connected resistor $= 4 \Omega$

resistance of connected lamp $= 20 \Omega$

Voltage of the battery $=\textrm 6 Volt$

We know that, ohm's law $Resistance = \frac{voltage}{Current}$  

(a) total resistance of the circuit $= \textrm{lamp resistance} +\textrm{series resistance}$

Total resistance of the circuit $= 20 + 4 = 24 \Omega$

(b) current flowing through the circuit $= \frac{Voltage}{Resistance}$

current flowing through the circuit $= \frac{6}{24} = \textrm{0.25 amp}$

(c) potential difference across the resistor$= Voltage \times Current$

potential difference across the resistor$= 4 \times 0.25 = \textrm{1 Volt}$

​(d) potential difference across the electric lamp$= Voltage \times Currents$

potential difference across the electric lamp$= 20 \times 0.25 = \textrm{5 Volt}$

Answer 2

Total Resistance of Circuit is $24 \Omega$

Explanation:

Given,

• The voltage of the Battery, V = 6 V

Resistance $R_1$ or Electric lamp = 20 $\Omega$

The resistance of Series Conductor $R_2$= 4 $\Omega$

• a.) Total Resistance in Circuit, $R_s = R_1 + R _ 2$

= 20 + 4 = 24 $\Omega$

• b.) Using Ohm's law,  

Current flowing through the circuit, I = $\frac {V}{R_s}$

= $\frac {6\V}{24\ \Omega}$

= 0.25  A

• c.) Potential Difference across the electric lamp,

$V_1 = IR_1 = 0.25 A \times 20\ \Omega$

= 5 V

• d.)  Potential Difference across the Conductor,

$V_2 = IR_2 = 0.25 A \times 4\ \Omega$

= 1 V