Question

consider the circuit shown in (figure 1). the current through the 6.00 ω resistor is 4.00 a, in the direction shown.

part a
what is the current through the 25.0 ω resistor?
express your answer with the appropriate units.

part b
what is the current through the 20.0 ω resistor?
express your answer with the appropriate units.
i20.0 = 9.95 a

Explanation:

Step1: Calculate the voltage across the \(6.00\ \Omega\) resistor

According to Ohm's law \(V = IR\). Given \(I = 4.00\ A\) and \(R=6.00\ \Omega\), we have \(V_{6}=4.00\times6.00 = 24.0\ V\).

Step2: Calculate the current through the \(8.00\ \Omega\) resistor

Since the \(8.00\ \Omega\) and \(6.00\ \Omega\) resistors are in parallel, the voltage across the \(8.00\ \Omega\) resistor \(V_{8}=V_{6} = 24.0\ V\). Using Ohm's law \(I=\frac{V}{R}\), for \(R = 8.00\ \Omega\), \(I_{8}=\frac{24.0}{8.00}=3.00\ A\).

Step3: Calculate the current through the \(25.0\ \Omega\) resistor (Part A)

The current through the \(25.0\ \Omega\) resistor is the sum of the currents through the \(6.00\ \Omega\) and \(8.00\ \Omega\) resistors (using the junction - rule). So \(I_{25}=4.00 + 3.00=7.00\ A\).

Step4: Calculate the voltage across the parallel - combination (of \(6.00\ \Omega\), \(8.00\ \Omega\) and \(25.0\ \Omega\))

The voltage across the parallel - combination \(V_{parallel}=I_{25}\times25.0+24.0\). Substituting \(I_{25} = 7.00\ A\), we get \(V_{parallel}=7.00\times25.0+24.0=175 + 24=199\ V\).

Step5: Calculate the current through the \(20.0\ \Omega\) resistor (Part B)

Since the \(20.0\ \Omega\) resistor is in parallel with the combination of \(6.00\ \Omega\), \(8.00\ \Omega\) and \(25.0\ \Omega\), the voltage across the \(20.0\ \Omega\) resistor \(V_{20}=V_{parallel}=199\ V\). Using Ohm's law \(I=\frac{V}{R}\), with \(R = 20.0\ \Omega\), \(I_{20}=\frac{199}{20.0}=9.95\ A\).

Answer:

Part A: The current through the \(25.0\ \Omega\) resistor \(I_{25.0}=7.00\ A\).
Part B: The current through the \(20.0\ \Omega\) resistor \(I_{20.0}=9.95\ A\).