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 b
what is the current through the 20.0 ω resistor?
express your answer with the appropriate units.

Explanation:

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

Use Ohm's law \(V = IR\). Here \(I = 4.00\ A\) and \(R=6.00\ \Omega\).
\(V_{6}=4.00\times6.00 = 24.0\ V\)

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

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

Step3: Calculate current through \(25.0\ \Omega\) resistor

Using Kirchhoff's current law at the junction, current through \(25.0\ \Omega\) resistor \(I_{25}=I_{8}+I_{6}\). \(I_{25}=3.00 + 4.00=7.00\ A\)

Step4: Calculate voltage across \(25.0\ \Omega\) resistor

Using Ohm's law \(V = IR\), \(V_{25}=7.00\times25.0=175\ V\)

Step5: Calculate total voltage across the parallel - branch

The voltage across the parallel - branch (where \(20.0\ \Omega\) is connected) is \(V = V_{25}+V_{6}=175 + 24.0=199\ V\)

Step6: Calculate current through \(20.0\ \Omega\) resistor

Using Ohm's law \(I=\frac{V}{R}\), with \(V = 199\ V\) and \(R = 20.0\ \Omega\), \(I_{20}=\frac{199}{20.0}=9.95\ A\)

Answer:

\(9.95\ A\)