A series circuit consist of 6 ohms resistor,ammeter and battery. The current in the circuit is 0.5A but drops to 0.3A when another resistor of 8ohms is added in series. Find the battery e.m.f and its internal resistance?
E = 0.5*r + 0.5*6
Eq1: E = 0.5r + 3.
E = 0.3*r + 0.3*(6+8).
Eq2: E = 0.3r + 4.2.
E = 0.5r + 3 = 0.3r + 4.2.
0.5r +3 = 0.3r + 4.2.
0.2r = 1.2.
r = 6 Ohms = Internal resistance.
E=0.5r + 0.5*R = 0.5*6 + 0.5*6 = 6 Volts
To find the battery electromotive force (e.m.f) and its internal resistance, we can use the information about the current in the circuit and the change in current when a resistor is added.
Let's analyze the circuit step by step:
1. Initial circuit: The circuit consists of a 6-ohm resistor, an ammeter (which measures the current), and the battery.
- The current in the circuit is given as 0.5A.
2. Circuit with an additional resistor: When another 8-ohm resistor is added in series with the initial components, the current in the circuit drops to 0.3A.
- So, the total resistance in the circuit can be calculated by using Ohm's Law: R = V / I.
- For the initial circuit, the total resistance is 6 ohms because there is only one resistor.
- For the modified circuit, the total resistance is the sum of the initial resistance (6 ohms) and the added resistance (8 ohms), which equals 14 ohms (6 ohms + 8 ohms).
3. Determining the battery e.m.f and internal resistance: We can use two equations to solve for the battery e.m.f (E) and internal resistance (r) of the circuit.
a) Kirchhoff's Voltage Law (KVL): The sum of the voltage drops across all components in a series circuit is equal to the e.m.f of the battery.
- In this circuit, the voltage drops across the resistors are V1 = 6I and V2 = 8I, where I is the current in the circuit.
- The voltage drop across the battery will then be V_bat = V1 + V2.
b) Ohm's Law: The voltage across a resistor is equal to the product of its resistance and the current passing through it.
- For the initial circuit, V1 (voltage across the 6-ohm resistor) is 6I.
- For the modified circuit, V2 (voltage across the 8-ohm resistor) is 8I.
Using the above information, let's form equations:
Equation 1: V_bat = V1 + V2
Equation 2: V1 = 6I
Equation 3: V2 = 8I
We can substitute equations 2 and 3 into equation 1 to obtain an equation solely in terms of I:
V_bat = 6I + 8I
Now substitute the known values of current in the circuit:
0.5A = 6 * 0.5A + 8 * 0.5A
Solving the equation:
0.5A = 3A + 4A
0.5A = 7A
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