A battery of emf 12v nd internal resistance (r) 1.5votls is connected to an external resistance (R) of 3ohms. (1.) Cal, the current flowing in the circuit. (2,) the p.d draw across the 3ohms resitor and the internal resirtance. (3) determine Voltage of the battery and comment of the result. (4) the reading of a volt metre connected across the batter therminal in the open circuit
The current is 2.67 amps
Potential drop across the 3 ohms is 8.01
Potential drop across internal resistance is 4.005
A battery of emf 12 volts and internal resistance r = 1.5 ohms is connected to an external resistance 2 ohms. Calculate (a) the current flowing in the circuit (b) the p.d drop across 2 ohms
1. Well, to find the current, we can use Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R). So, in this case, the total resistance (R_total) would be the sum of the external resistance (R) and the internal resistance (r). Therefore, R_total = R + r = 3Ω + 1.5Ω = 4.5Ω. Now we can use the formula I = V / R_total to calculate the current flowing in the circuit:
I = 12V / 4.5Ω ≈ 2.67A
2. To find the potential difference (p.d.) across the 3Ω resistor and the internal resistance, we can use Ohm's Law again. Since the internal resistance and the external resistance are in series, the total current flowing through them is the same. Therefore, the p.d. across each resistor can be calculated separately.
p.d. across the 3Ω resistor (V_R) = I * R = 2.67A * 3Ω = 8.01V
p.d. across the internal resistance (V_r) = I * r = 2.67A * 1.5Ω = 4.005V
3. The voltage of the battery is given as 12V. This is the electromotive force (emf) of the battery, which represents its maximum possible voltage without any current flowing. It's important to note that the actual voltage across the battery terminals will be slightly less than the emf due to the internal resistance. In this case, the voltage across the battery terminals (V_batt) can be found by subtracting the voltage across the internal resistance (V_r) from the emf:
V_batt = emf - V_r = 12V - 4.005V ≈ 7.995V
As for commenting on the result, well, it seems like our battery is losing some voltage due to its internal resistance. It's like losing money in hidden fees. Not cool, battery, not cool.
4. If a voltmeter is connected across the battery terminals in an open circuit, it means that no current is flowing. Therefore, there will be no drop in voltage across the internal resistance. As a result, the voltmeter will read the same voltage as the battery's emf, which is 12V in this case. So, it's a perfect match! Just like when you're dating and both you and your partner have a good sense of humor.
To solve this problem, we can apply Ohm's Law and Kirchhoff's Laws to determine the current, potential difference, and voltage of the battery. Here's how:
1. To calculate the current flowing in the circuit, we can use Ohm's Law: I = V / R, where I is the current, V is the potential difference, and R is the resistance. In this case, the external resistance is given as 3 ohms. So, the current can be calculated as I = 12V / 3Ω = 4A.
2. To find the potential difference across the 3-ohm resistor and the internal resistance, we can calculate the voltage drops separately. The potential difference across the 3-ohm resistor can be found using Ohm's Law: V = I * R. So, V = 4A * 3Ω = 12V. The potential difference across the internal resistance can be calculated similarly: V = I * r = 4A * 1.5Ω = 6V.
3. To determine the voltage of the battery, we can use Kirchhoff's Voltage Law (KVL) which states that the sum of all voltage drops in a closed loop is equal to the sum of all electromotive forces (emf). In this case, the voltage across the external resistor will be the sum of the battery voltage and the voltage drop across the internal resistance. So, V = E + V_internal. Substitute the values given: 12V = E + 6V. Solving, we find that the voltage of the battery (E) is 6V.
Comment: The voltage of the battery (E) is smaller than the emf (12V) due to the voltage drop across the internal resistance. This decrease in voltage is expected as there is always some energy loss within a real battery due to its internal resistance.
4. When a voltmeter is connected across the battery terminals in an open circuit, no current flows through the circuit. According to Ohm's Law, if there is no current (I = 0), the potential difference (V) across the battery terminals will be equal to the emf of the battery. Therefore, the reading on the voltmeter will show the emf of the battery, which in this case is 12V.
I hope this explanation helps you understand how to solve the given problem.