Question

1. in the circuit shown in fig. 7 find (a) the current in resistor ( r ); (b) the resistance ( r ); (c) the unknown emf ( mathcal{e} ). (d) if the circuit is broken at point ( x ), what is the current in resistor ( r )?

Explanation:

Step1: Find current in resistor \( R \) (part a)

Using junction rule (conservation of charge), the current in \( R \) is \( I = 6.00\,\text{A}-4.00\,\text{A}=2.00\,\text{A}\)

Step2: Find resistance \( R \) (part b)

Using loop rule (conservation of energy), for the loop with \( 28.0\,\text{V}\) battery, \( R \), \( 3.00\,\Omega\) resistor: \( 28.0\,\text{V}=I\times R + 6.00\,\text{A}\times3.00\,\Omega\). Substitute \( I = 2.00\,\text{A}\), then \( 28.0=2R + 18\), \( 2R=28 - 18=10\), \( R = 5.00\,\Omega\)

Step3: Find unknown emf \( \mathcal{E} \) (part c)

Using loop rule for the loop with \( \mathcal{E}\), \( 4.00\,\text{A}\) current and \( 6.00\,\Omega\) resistor: \( \mathcal{E}=4.00\,\text{A}\times6.00\,\Omega=24.0\,\text{V}\)

Step4: Find current in \( R \) when circuit is broken at \( x \) (part d)

When broken at \( x \), the current in the circuit is \( I'=\frac{28.0\,\text{V}}{R + 3.00\,\Omega}\). Substitute \( R = 5.00\,\Omega\), \( I'=\frac{28}{5 + 3}=\frac{28}{8}=3.50\,\text{A}\)

Answer:

(a) \( 2.00\,\text{A}\)
(b) \( 5.00\,\Omega\)
(c) \( 24.0\,\text{V}\)
(d) \( 3.50\,\text{A}\)