Question

Consider the circuit shown in (Figure 1). Suppose that E = 15 V . include units with answers.

Part A: Find the current through the resistor a.

Part B: Find the potential difference across the resistor a. answer: 7.5 V

Part C: Find the current through the resistor b.

Part D: Find the potential difference across the resistor b.

Part E: Find the current through the resistor c.

Part F: Find the potential difference across the resistor c.

Part G: Find the current through the resistor d.

Part H: Find the potential difference across the resistor d.

Answer 1

Solution The resistances c and d are in series 

$$ \begin{aligned} R_{x} &=R_{e}+R_{d} \\ &=5 \Omega+5 \Omega \\ &=10 \Omega \end{aligned} $$

Thus, total resistance in series is \(10 \Omega\)

The following diagram represents the simplified circuit diagram

Resistance \(b\) and \(R_{x}\) in parallel \(\frac{1}{R_{p}}=\frac{1}{R_{s}}+\frac{1}{R_{s}}\)

\(=\frac{1}{10 \Omega}+\frac{1}{10 \Omega}\)

\(=\frac{1}{5}\)

\(R_{p}=5 \Omega\)

Thus, total resistance in parallel is \(5 \Omega\)

 

$$ \begin{array}{l} \text { The following diagram represents the simplified circuit diagram } \\ \qquad 15 \mathrm{v}=\mathrm{VM} \end{array} $$

 

Equivalent resistance in the circuit

$$ \begin{aligned} R_{e q_{0}} &=R_{a}+R_{p} \\ &=5 \Omega+5 \Omega \\ &=10 \Omega \end{aligned} $$

Therefore, Equivalent resistance in the circuit is \(10 \Omega\)

(a) Current through the resistance a is, By Ohm's law

\(\begin{aligned} I_{a} &=\frac{V}{R_{\mathrm{eq} u}} \\ &=\frac{15 \mathrm{~V}}{10} \\ &=1.5 \mathrm{~A} \end{aligned}\)

Therefore, current through resistance a is \(1.5 \mathrm{~A}\)

(b) Potential difference through resistance a 

\(\begin{aligned} V_{a} &=I_{a} R_{a} \\ &=1.5 \mathrm{~V} \times 5 \Omega \\ &=7.5 \mathrm{~V} \end{aligned}\)

Therefore, potential difference across a is \(7.5 \mathrm{~V}\)

 

(c) Current through the resistance b By \(O h m^{\prime}\) s law

\(\begin{aligned} I_{b} &=R_{\text {equ }} \times \frac{I_{a}}{R_{s}+R_{x}} \\ &=10 \Omega \times \frac{1.5 \mathrm{~A}}{10 \Omega+10 \Omega} \\ &=0.75 \mathrm{~A} \end{aligned}\)

Therefore, current through resistance b is \(0.75 \mathrm{~A}\)

 

(d)

Potential difference through resistance b

$$ \begin{aligned} V_{b} &=I_{b} R_{b} \\ &=0.75 \mathrm{~A} \times 10 \Omega \\ &=7.5 \mathrm{~V} \end{aligned} $$

Therefore, potential difference across b is \(\overline{7.5 \mathrm{~V}}\)

(e)

Current through the resistance By \(\mathrm{Ohm}^{\prime}\) s law

\(I_{e}=I_{a}-I_{b}\)

$$ \begin{array}{l} =1.5-0.75 \\ =0.75 \mathrm{~A} \end{array} $$

Therefore, current through resistance c is \(0.75 \mathrm{~A}\)

(f) Potential difference through resistance \(V_{e}=I_{e} R_{e}\)

$$ \begin{array}{l} =0.75 \mathrm{~A} \times 5 \Omega \\ =3.75 \mathrm{~V} \end{array} $$

Therefore, potential difference across c is \(3.75 \mathrm{~V}\)

 

(f) Potential difference through resistance \(V_{e}=I_{e} R_{e}\)

$$ \begin{array}{l} =0.75 \mathrm{~A} \times 5 \Omega \\ =3.75 \mathrm{~V} \end{array} $$

Therefore, potential difference across c is \(3.75 \mathrm{~V}\)

(g) Current through the resistance d By \(O h m^{\prime}\) s law

\(\begin{aligned} I_{d} &=I_{a}-I_{b} \\ &=1.5-0.75 \\ &=0.75 \mathrm{~A} \end{aligned}\)

Therefore, current through resistance d is \(0.75 \mathrm{~A}\)

 

(h) Potential difference through resistance d

$$ \begin{aligned} V_{d} &=I_{d} R_{d} \\ &=0.75 \mathrm{~A} \times 5 \Omega \\ &=3.75 \mathrm{~V} \end{aligned} $$

Therefore, potential difference across d is \(3.75 \mathrm{~V}\)