(a) State Ohm's law. Define the SI unit of electrical resistance. (b) An electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6 ext{ V}$ battery. Calculate: (i) the total resistance of the circuit, (ii) the current through the circuit, and (iii) the potential difference across the electric lamp and conductor.

Q: (a) State Ohm's law. Define the SI unit of electrical resistance. (b) An electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6\text{ V}$ battery. Calculate: (i) the total resistance of the circuit, (ii) the current through the circuit, and (iii) the potential difference across the electric lamp and conductor.

(a) Statement of Ohm's Law and SI Unit of Resistance Statement of Ohm's Law: Ohm's law states that the physical conditions (such as temperature) remaining unchanged, the electric current ($I$) flowing through a metallic wire is directly proportional to the potential difference ($V$) applied across its ends. Mathematically, $V \propto I$ or $V = IR$, where $R$ is a constant called the resistance of the conductor. Definition of SI Unit of Resistance: The SI unit of electric resistance is ohm ($?\Omega$). According to Ohm's law, $R = \frac{V}{I}$. If a potential difference of $1\text{ volt}$ is applied across the ends of a conductor and a current of $1\text{ ampere}$ flows through it, then the resistance of the conductor is said to be $1\text{ ohm}$. Therefore, $1\ \Omega = \frac{1\text{ V}}{1\text{ A}}$. (b) Numerical Problem Step-by-Step Solution Given Data: Resistance of the electric lamp, $R1 = 20\ \Omega$ Resistance of the conductor, $R2 = 4\ \Omega$ Voltage of the battery, $V = 6\text{ V}$ (i) Total resistance of the circuit ($Rs$): Since the lamp and conductor are connected in series, the total resistance is the sum of individual resistances. $$Rs = R1 + R2$$ $$Rs = 20\ \Omega + 4\ \Omega = 24\ \Omega$$ (ii) Current through the circuit ($I$): Using Ohm's law, the total current in the circuit is given by: $$I = \frac{V}{Rs}$$ $$I = \frac{6\text{ V}}{24\ \Omega} = 0.25\text{ A}$| (iii) Potential difference across the electric lamp ($V1$) and conductor ($V2$): Potential difference across the electric lamp ($V1$): $$V1 = I \times R1$$ $$V1 = 0.25\text{ A} \times 20\ \Omega = 5.0\text{ V}$$ Potential difference across the conductor ($V2$): $$V2 = I \times R2$$ $$V2 = 0.25\text{ A} \times 4\ \Omega = 1.0\text{ V}$$

Question

(a) State Ohm's law. Define the SI unit of electrical resistance. (b) An electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6	ext{ V}$ battery. Calculate: (i) the total resistance of the circuit, (ii) the current through the circuit, and (iii) the potential difference across the electric lamp and conductor.

Accepted Answer

(a) Statement of Ohm's Law and SI Unit of Resistance

Statement of Ohm's Law: Ohm's law states that the physical conditions (such as temperature) remaining unchanged, the electric current ($I$) flowing through a metallic wire is directly proportional to the potential difference ($V$) applied across its ends. Mathematically, $V \propto I$ or $V = IR$, where $R$ is a constant called the resistance of the conductor.
Definition of SI Unit of Resistance: The SI unit of electric resistance is ohm ($?\Omega$). According to Ohm's law, $R = \frac{V}{I}$. If a potential difference of $1\text{ volt}$ is applied across the ends of a conductor and a current of $1\text{ ampere}$ flows through it, then the resistance of the conductor is said to be $1\text{ ohm}$. Therefore, $1\ \Omega = \frac{1\text{ V}}{1\text{ A}}$.

(b) Numerical Problem Step-by-Step Solution

Given Data:
- Resistance of the electric lamp, $R_1 = 20\ \Omega$
- Resistance of the conductor, $R_2 = 4\ \Omega$
- Voltage of the battery, $V = 6\text{ V}$
(i) Total resistance of the circuit ($R_s$): Since the lamp and conductor are connected in series, the total resistance is the sum of individual resistances. $$R_s = R_1 + R_2$$ $$R_s = 20\ \Omega + 4\ \Omega = 24\ \Omega$$
(ii) Current through the circuit ($I$): Using Ohm's law, the total current in the circuit is given by: $$I = \frac{V}{R_s}$$ $$I = \frac{6\text{ V}}{24\ \Omega} = 0.25\text{ A}$|
(iii) Potential difference across the electric lamp ($V_1$) and conductor ($V_2$):
- Potential difference across the electric lamp ($V_1$): $$V_1 = I \times R_1$$ $$V_1 = 0.25\text{ A} \times 20\ \Omega = 5.0\text{ V}$$
- Potential difference across the conductor ($V_2$): $$V_2 = I \times R_2$$ $$V_2 = 0.25\text{ A} \times 4\ \Omega = 1.0\text{ V}$$