JEE Advanced

• For the first problem, we can use Ohm’s law to calculate the current flowing in the circuit. The total resistance of the circuit is 10 Ω + 2 Ω = 12 Ω. Therefore, the current flowing in the circuit is I = V/R = 12 V / 12 Ω = 1 A.
• For the second problem, we can first calculate the equivalent resistance of the two 2 Ω resistors in parallel. The equivalent resistance is given by 1/R = 1/R1 + 1/R2, where R1 and R2 are the resistances of the two resistors. Substituting R1 = R2 = 2 Ω, we get 1/R = 1/2 + 1/2 = 1. Therefore, the equivalent resistance is R = 1 Ω. Now, we can use Ohm’s law to calculate the total current flowing in the circuit. The total resistance of the circuit is 1 Ω + 1 Ω = 2 Ω. Therefore, the total current flowing in the circuit is I = V/R = 6 V / 2 Ω = 3 A.
• For the third problem, we can simply add the voltages of the two batteries to get the total voltage of the combination. The total voltage is 2 V + 4 V = 6 V. Now, we can use Ohm’s law to calculate the current flowing in the circuit. The total resistance of the circuit is 4 Ω. Therefore, the current flowing in the circuit is I = V/R = 6 V / 4 Ω = 1.5 A.
• For the fourth problem, we can first calculate the equivalent resistance of the 10 Ω resistor and the 5 Ω resistor in parallel. The equivalent resistance is given by 1/R = 1/R1 + 1/R2, where R1 and R2 are the resistances of the two resistors. Substituting R1 = 10 Ω and R2 = 5 Ω, we get 1/R = 1/10 + 1/5 = 3/10. Therefore, the equivalent resistance is R = 10/3 Ω. Now, we can use Ohm’s law to calculate the current flowing through the 5 Ω resistor. The current flowing through the 5 Ω resistor is I = V/R = 12 V / (5 Ω + 10/3 Ω) ≈ 0.86 A.

CBSE Board

• For the first problem, we can use Ohm’s law to calculate the current flowing in the circuit. The total resistance of the circuit is 5 Ω + 1 Ω = 6 Ω. Therefore, the current flowing in the circuit is I = V/R = 2 V / 6 Ω = 1/3 A.
• For the second problem, we can first calculate the equivalent resistance of the three 2 Ω resistors in parallel. The equivalent resistance is given by 1/R = 1/R1 + 1/R2 + 1/R3, where R1, R2, and R3 are the resistances of the three resistors. Substituting R1 = R2 = R3 = 2 Ω, we get 1/R = 1/2 + 1/2 + 1/2 = 3/2. Therefore, the equivalent resistance is R = 2/3 Ω. Now, we can use Ohm’s law to calculate the total current flowing in the circuit. The total resistance of the circuit is 2/3 Ω + 2 Ω = 8/3 Ω. Therefore, the total current flowing in the circuit is I = V/R = 6 V / 8/3 Ω = 9/4 A.
• For the third problem, we can simply add the voltages of the two batteries to get the total voltage of the combination. The total voltage is 1.5 V + 3 V = 4.5 V. Now, we can use Ohm’s law to calculate the current flowing in the circuit. The total resistance of the circuit is 4 Ω. Therefore, the current flowing in the circuit is I = V/R = 4.5 V / 4 Ω = 1.125 A.
• For the fourth problem, we can first calculate the equivalent resistance of the 5 Ω resistor and the 3 Ω resistor in parallel. The equivalent resistance is given by 1/R = 1/R1 + 1/R2, where R1 and R2 are the resistances of the two resistors. Substituting R1 = 5 Ω and R2 = 3 Ω, we get 1/R = 1/5 + 1/3 = 8/15. Therefore, the equivalent resistance is R = 15/8 Ω. Now, we can use Ohm’s law to calculate the current flowing through the 3 Ω resistor. The current flowing through the 3 Ω resistor is I = V/R = 9 V / (3 Ω + 15/8 Ω) ≈ 2.12 A.