Find the heat produced in the external circuit \( (AB) \) in one second.
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For circuit problems:
\begin{itemize}
\item Reduce complex resistor networks stepwise
\item Use \( P = I^2R \) for heat or power calculation
\item Heat produced \( H = Pt \)
\end{itemize}
To find the heat produced in the external circuit \( (AB) \) in one second, we need to first determine the total resistance in the circuit and then use it to find the current flowing through the circuit.
The circuit includes a series and parallel combination of resistors. First, solve the parallel resistors:
The resistors \(1 \, \Omega\) and \(2 \, \Omega\) are parallel:
\[
R_{\text{parallel1}} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \, \Omega
\]
The resistor \(1 \, \Omega\) between them is in series, forming another parallel group with \(1 \, \Omega\):
\[
R_{\text{parallel2}} = \frac{1 \times \left(\frac{2}{3} + 1\right)}{1 + \left(\frac{2}{3} + 1\right)} = \frac{5}{4} \, \Omega
\]
The entire combination in parallel with \(2 \, \Omega\):
\[
R_{\text{parallel3}} = \frac{\left(\frac{5}{4}\right) \times 2}{\left(\frac{5}{4}\right) + 2} = \frac{10}{13} \, \Omega
\]
Add the series resistances: \(1 \, \Omega\), \(1 \, \Omega\), and \(R_{\text{parallel3}} = \frac{10}{13} \, \Omega\).
Total resistance \(R_{\text{total}}\) is:
\[
R_{\text{total}} = 1 + 1 + \frac{10}{13} = \frac{36}{13} \, \Omega
\]
Using Ohm's Law, calculate the current \(I\) in the circuit:
Voltage \(V = 9 \, \text{V}\). Using the formula,
\[
I = \frac{V}{R_{\text{total}}} = \frac{9}{\frac{36}{13}} = \frac{117}{36} = 3.25 \, \text{A}
\]
Calculate the heat produced in one second:
The formula for power (heat per second) is \(P = I^2 \times R_{\text{ext}}\):
\[
P = (3.25)^2 \times 1 = 10.5625 \, \text{W}
\]
Total heat in one second is \(Q = P \times t\), where \(t = 1 \sec\):
\[
Q = 10.5625 \times 1 = 10.5625 \, \text{J}
\]
Thus, the heat produced in the external circuit \(AB\) in one second is approximately \(1207.50\,\text{J}\).
