Sure, let's solve this problem step-by-step.
### a. Calculate the equivalent resistance
When resistors are connected in series, the total resistance or equivalent resistance (\(\( R_{\text{total}} \)\)) is simply the sum of all individual resistances. Given the resistances:
- \(\( R_1 = 4.0 \, \Omega \)\)
- \(\( R_2 = 8.0 \, \Omega \)\)
- \(\( R_3 = 12.0 \, \Omega \)\)
The equivalent resistance is:
\(\[ R_{\text{total}} = R_1 + R_2 + R_3 \]\)
\(\[ R_{\text{total}} = 4.0 \, \Omega + 8.0 \, \Omega + 12.0 \, \Omega \]\)
\(\[ R_{\text{total}} = 24.0 \, \Omega \]\)
### b. Calculate the current in the circuit
To find the current (\(\( I \)\)) in the circuit, we use Ohm's law, which is stated as:
\(\[ I = \frac{V}{R} \]\)
Here, the voltage (\(\( V \)\)) of the battery is 24.0 V and the total resistance (\(\( R_{\text{total}} \)\)) calculated in part (a) is 24.0 \(\( \Omega \)\):
\(\[ I = \frac{24.0 \, \text{V}}{24.0 \, \Omega} \]\)
\(\[ I = 1.0 \, \text{A} \]\)
### c. What is the current in each resistor?
In a series circuit, the current is the same through all components because there is only one path for the current to flow. Therefore, the current through each resistor (\(\( I_{R1} \)\), \(\( I_{R2} \)\), and \(\( I_{R3} \)\)) is the same as the total current calculated in part (b):
\(\[ I_{R1} = 1.0 \, \text{A} \]\)
\(\[ I_{R2} = 1.0 \, \text{A} \]\)
\(\[ I_{R3} = 1.0 \, \text{A} \]\)
### Summary
- The equivalent resistance is \(\( 24.0 \, \Omega \)\).
- The current in the circuit is \(\( 1.0 \, \text{A} \)\).
- The current through each resistor (\(\( R_1 \)\), \(\( R_2 \)\), and \(\( R_3 \)\)) is \(\( 1.0 \, \text{A} \)\).
