How much heat is produced in the 10-Ω resistor in 5.0 s when ε = 18 V, given the circuit has a 12-Ω and a 15-Ω resistor in series with the 10-Ω resistor?

Steps to Solve

1. Calculate the total resistance in the series circuit

In a series circuit, the total resistance ($R_{total}$) is the sum of individual resistances:

$R_{total} = R_1 + R_2 + R_3$

$R_{total} = 10 , \Omega + 12 , \Omega + 15 , \Omega = 37 , \Omega$

2. Calculate the current flowing through the circuit using Ohm's Law

Ohm's Law states that $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance. We can rearrange this to solve for the current $I$:

$I = \frac{V}{R_{total}}$

$I = \frac{18 , V}{37 , \Omega} \approx 0.486 , A$

3. Calculate the power dissipated by the 10-Ω resistor

The power $P$ dissipated by a resistor is given by $P = I^2R$. In this case, we want to find the power dissipated by the 10-Ω resistor ($R = 10 , \Omega$).

$P = (0.486 , A)^2 \times 10 , \Omega$

$P \approx 2.36 , W$

4. Calculate the heat produced in the 10-Ω resistor over 5.0 seconds

Heat ($H$) is the energy dissipated over time, and can be calculated using the formula $H = P \times t$, where $P$ is power and $t$ is time.

$H = 2.36 , W \times 5.0 , s$

$H \approx 11.8 , J$