Answer :
To find the internal resistance \( r \) of each cell, let's analyze the situation step-by-step.
1. Understanding the Circuit:
- We have two identical cells, each with an electromotive force (EMF) of 1.5 V.
- The cells are connected in parallel, so the total EMF is still 1.5 V.
- The current passing through the external resistor is 0.6 A.
- The value of the external resistor is 2.0 Ω.
2. Applying Ohm's Law:
- Ohm's Law states: \( V = I \cdot R \)
- Here, \( V \) is the total voltage (1.5 V), \( I \) is the current (0.6 A), and \( R \) is the total resistance in the circuit.
- The total resistance in the circuit is the sum of the external resistor and the internal resistance of the cells (since internal resistances of cells in parallel should be equivalent to a single internal resistance \( r \), as per the question).
3. Setting Up the Equation:
- From Ohm's Law: \( V = I \cdot (R_{\text{external}} + R_{\text{internal}}) \)
- Plugging in the values: \( 1.5 = 0.6 \cdot (2.0 + r) \)
4. Solving for Internal Resistance (\( r \)):
- Rearrange the equation to isolate \( r \):
[tex]\[ 1.5 = 0.6 \cdot (2.0 + r) \][/tex]
- Distribute the current across the resistances:
[tex]\[ 1.5 = 1.2 + 0.6r \][/tex]
- Subtract 1.2 from both sides of the equation:
[tex]\[ 1.5 - 1.2 = 0.6r \][/tex]
- Simplify:
[tex]\[ 0.3 = 0.6r \][/tex]
- Divide both sides by 0.6 to solve for \( r \):
[tex]\[ r = \frac{0.3}{0.6} = 0.5 \][/tex]
So, the value of the internal resistance [tex]\( r \)[/tex] of each cell is [tex]\( 0.5 \)[/tex] Ω.
1. Understanding the Circuit:
- We have two identical cells, each with an electromotive force (EMF) of 1.5 V.
- The cells are connected in parallel, so the total EMF is still 1.5 V.
- The current passing through the external resistor is 0.6 A.
- The value of the external resistor is 2.0 Ω.
2. Applying Ohm's Law:
- Ohm's Law states: \( V = I \cdot R \)
- Here, \( V \) is the total voltage (1.5 V), \( I \) is the current (0.6 A), and \( R \) is the total resistance in the circuit.
- The total resistance in the circuit is the sum of the external resistor and the internal resistance of the cells (since internal resistances of cells in parallel should be equivalent to a single internal resistance \( r \), as per the question).
3. Setting Up the Equation:
- From Ohm's Law: \( V = I \cdot (R_{\text{external}} + R_{\text{internal}}) \)
- Plugging in the values: \( 1.5 = 0.6 \cdot (2.0 + r) \)
4. Solving for Internal Resistance (\( r \)):
- Rearrange the equation to isolate \( r \):
[tex]\[ 1.5 = 0.6 \cdot (2.0 + r) \][/tex]
- Distribute the current across the resistances:
[tex]\[ 1.5 = 1.2 + 0.6r \][/tex]
- Subtract 1.2 from both sides of the equation:
[tex]\[ 1.5 - 1.2 = 0.6r \][/tex]
- Simplify:
[tex]\[ 0.3 = 0.6r \][/tex]
- Divide both sides by 0.6 to solve for \( r \):
[tex]\[ r = \frac{0.3}{0.6} = 0.5 \][/tex]
So, the value of the internal resistance [tex]\( r \)[/tex] of each cell is [tex]\( 0.5 \)[/tex] Ω.
