Sure! Let's go through the steps required to answer each part of the question:
### Given:
- Voltage of the battery ([tex]\(V_{\text{battery}}\)[/tex]) = 20 V
- Internal resistance ([tex]\(R_{\text{internal}}\)[/tex]) = 5 Ω
- Resistor ([tex]\(R_{\text{resistor}}\)[/tex]) = 20 Ω
### Part (a): Electric Power (P)
1. Calculate the total resistance in the circuit:
[tex]\[ R_{\text{total}} = R_{\text{internal}} + R_{\text{resistor}} \][/tex]
[tex]\[ R_{\text{total}} = 5 \, \Omega + 20 \, \Omega = 25 \, \Omega \][/tex]
2. Calculate the current using Ohm's Law:
[tex]\[ I = \frac{V_{\text{battery}}}{R_{\text{total}}} \][/tex]
[tex]\[ I = \frac{20 \, \text{V}}{25 \, \Omega} = 0.8 \, \text{A} \][/tex]
3. Calculate the electric power using the formula [tex]\( P = V \times I \)[/tex]:
[tex]\[ P = 20 \, \text{V} \times 0.8 \, \text{A} = 16 \, \text{W} \][/tex]
### Part (b): Terminal Voltage ([tex]\(V_{\text{terminal}}\)[/tex])
1. Use the current calculated above and the resistor value to find the terminal voltage:
[tex]\[ V_{\text{terminal}} = I \times R_{\text{resistor}} \][/tex]
[tex]\[ V_{\text{terminal}} = 0.8 \, \text{A} \times 20 \, \Omega = 16 \, \text{V} \][/tex]
### Part (c): Lost Voltage ([tex]\(V_{ \text{lost}}\)[/tex])
1. Use the current calculated above and the internal resistance to find the lost voltage:
[tex]\[ V_{\text{lost}} = I \times R_{\text{internal}} \][/tex]
[tex]\[ V_{\text{lost}} = 0.8 \, \text{A} \times 5 \, \Omega = 4 \, \text{V} \][/tex]
### Final Answer:
- (a) Electric Power (P): [tex]\(16 \, \text{W}\)[/tex]
- (b) Terminal Voltage ([tex]\(V_{\text{terminal}}\)[/tex]): [tex]\(16 \, \text{V}\)[/tex]
- (c) Lost Voltage ([tex]\(V_{\text{lost}}\)[/tex]): [tex]\(4 \, \text{V}\)[/tex]
