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 ([tex]\( R_{\text{total}} \)[/tex]) is simply the sum of all individual resistances. Given the resistances:
- [tex]\( R_1 = 4.0 \, \Omega \)[/tex]
- [tex]\( R_2 = 8.0 \, \Omega \)[/tex]
- [tex]\( R_3 = 12.0 \, \Omega \)[/tex]
The equivalent resistance is:
[tex]\[ R_{\text{total}} = R_1 + R_2 + R_3 \][/tex]
[tex]\[ R_{\text{total}} = 4.0 \, \Omega + 8.0 \, \Omega + 12.0 \, \Omega \][/tex]
[tex]\[ R_{\text{total}} = 24.0 \, \Omega \][/tex]
### b. Calculate the current in the circuit
To find the current ([tex]\( I \)[/tex]) in the circuit, we use Ohm's law, which is stated as:
[tex]\[ I = \frac{V}{R} \][/tex]
Here, the voltage ([tex]\( V \)[/tex]) of the battery is 24.0 V and the total resistance ([tex]\( R_{\text{total}} \)[/tex]) calculated in part (a) is 24.0 [tex]\( \Omega \)[/tex]:
[tex]\[ I = \frac{24.0 \, \text{V}}{24.0 \, \Omega} \][/tex]
[tex]\[ I = 1.0 \, \text{A} \][/tex]
### 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 ([tex]\( I_{R1} \)[/tex], [tex]\( I_{R2} \)[/tex], and [tex]\( I_{R3} \)[/tex]) is the same as the total current calculated in part (b):
[tex]\[ I_{R1} = 1.0 \, \text{A} \][/tex]
[tex]\[ I_{R2} = 1.0 \, \text{A} \][/tex]
[tex]\[ I_{R3} = 1.0 \, \text{A} \][/tex]
### Summary
- The equivalent resistance is [tex]\( 24.0 \, \Omega \)[/tex].
- The current in the circuit is [tex]\( 1.0 \, \text{A} \)[/tex].
- The current through each resistor ([tex]\( R_1 \)[/tex], [tex]\( R_2 \)[/tex], and [tex]\( R_3 \)[/tex]) is [tex]\( 1.0 \, \text{A} \)[/tex].
