Question

Figure 1 illustrate a circuit consists of series and parallel resistors. By using Ohm’s Law, determine following parameters:i) total resistance of the circuit, Reqii) current iiii) voltage Vxiv) power dissipated at R5

Answer 1

i) Proceed as follow:

Sum R6, R7 and R8 in parallel:

$\begin{gathered} \frac{1}{R^{\prime}}=\frac{1}{R6}+\frac{1}{R7}+\frac{1}{R8} \\ \frac{1}{R^{\prime}}=\frac{1}{2000\Omega}+\frac{1}{550\Omega}+\frac{1}{1000\Omega} \\ R^{\prime}\approx301.37\Omega \end{gathered}$

Next, sum R3 and R4 in parallel;

$\begin{gathered} \frac{1}{R^{\doubleprime}}=\frac{1}{R3}+\frac{1}{R4} \\ \frac{1}{R^{\doubleprime}}=\frac{1}{900\Omega}+\frac{1}{1500\Omega} \\ R^{\doubleprime}=562.5\Omega \end{gathered}$

Then, you obtain a circuit with only resistances in series. Sum R1, R2, R'', R5, R' and R9 in series:

$\begin{gathered} R_{\text{tot}}=R1+R2+R^{\doubleprime}+R5+R^{\prime}+R9 \\ R_{\text{tot}}=550\Omega+1000\Omega+562.5\Omega+220\Omega+301.37\Omega+200\Omega \\ R_{\text{tot}}=2833.87\Omega \end{gathered}$

The previous result is the total resistance of the circuit

ii) Current I is given by:

$i=\frac{V}{R_{\text{tot}}}=\frac{80V}{2833.87\Omega}=0.028A$

The current is 0.028A

iii) Voltage at Vx is:

$V_x=i\cdot R5=(0.028A)(220\Omega)=6.16V$

The voltage is 6.16V

iv) The power dissipated at R5 is:

$P=i^2R5=(0.028A)^2(220\Omega)=0.17W$

Hence, the power dissiapted at R5 is approximately 0.17W