Certainly! Let's go step-by-step through solving this problem involving resistors in parallel and a battery.
### a) Find the current through each branch of the circuit.
The voltage across each resistor in a parallel circuit is equal to the voltage of the battery. We are given:
- Voltage [tex]\( V = 30.0 \)[/tex] volts
- Each resistor [tex]\( R = 15.0 \)[/tex] ohms
To find the current through each resistor, we use Ohm’s law, which states [tex]\( I = \frac{V}{R} \)[/tex]:
[tex]\[
I = \frac{30.0 \, \text{V}}{15.0 \, \text{ohms}} = 2.0 \, \text{A}
\][/tex]
So, the current through each branch (resistor) is [tex]\( 2.0 \)[/tex] amperes.
### b) Find the equivalent resistance of the circuit.
When resistors are connected in parallel, the reciprocal of the equivalent resistance [tex]\( R_{\text{eq}} \)[/tex] is the sum of the reciprocals of the individual resistances. There are three resistors, each with a resistance [tex]\( R = 15.0 \)[/tex] ohms. The formula for the equivalent resistance in a parallel circuit is:
[tex]\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\][/tex]
Since all resistors are the same:
[tex]\[
\frac{1}{R_{\text{eq}}} = \frac{1}{15.0 \, \text{ohms}} + \frac{1}{15.0 \, \text{ohms}} + \frac{1}{15.0 \, \text{ohms}} = \frac{3}{15.0 \, \text{ohms}} = \frac{1}{5.0 \, \text{ohms}}
\][/tex]
Thus, the equivalent resistance [tex]\( R_{\text{eq}} \)[/tex] is:
[tex]\[
R_{\text{eq}} = 5.0 \, \text{ohms}
\][/tex]
### c) Find the current through the battery.
To find the total current supplied by the battery, we use Ohm’s law again, but this time with the equivalent resistance of the circuit:
[tex]\[
I_{\text{total}} = \frac{V}{R_{\text{eq}}}
\][/tex]
Given the battery voltage [tex]\( V = 30.0 \)[/tex] volts and the equivalent resistance [tex]\( R_{\text{eq}} = 5.0 \)[/tex] ohms:
[tex]\[
I_{\text{total}} = \frac{30.0 \, \text{V}}{5.0 \, \text{ohms}} = 6.0 \, \text{A}
\][/tex]
So, the total current through the battery is [tex]\( 6.0 \, \text{A} \)[/tex].
### Summary:
a) The current through each branch of the circuit is [tex]\( 2.0 \)[/tex] amperes.
b) The equivalent resistance of the circuit is [tex]\( 5.0 \)[/tex] ohms.
c) The total current through the battery is [tex]\( 6.0 \)[/tex] amperes.
