Question Please explain 2. Consider electric circuits below and find: a) total resistance ANSWER: b) potential difference across the resistor R4​ ANSWER: c) current through resistor R2​ ANSWER: d) total power dissipated by the circuit ANSWER: e) power dissipated by R3​ ANSWER: f) indicate on the drawing where and how would you connect instrument to measure potential difference ΔV across resistor R1​ and current I through R3​. Indicate the name of each instrument (A,V) on it. R1​=12Ω;R2​=16.0Ω;R3​=24Ω;R4​=18Ω,R5​=12Ω. The voltage of the source is 12 V.

Question 32 (a) Rasistors R_{1} \& R_{2} are in parallel so their equivalent revistanceR_{6}=\frac{R_{1} \times R_{2}}{R_{1}+R_{2}}=\frac{12 \times 16}{12+16}=6.86 \OmegaResistors R_{3}, R_{4} \& R_{5} one in parallel so their equivalent reristance\begin{aligned}\frac{1}{R_{7}} & =\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}} \\& =\frac{1}{24}+\frac{1}{18}+\frac{1}{12} \\\text { or } R_{7} & =5.54 \Omega\end{aligned}Resistars R_{6} \& R_{7} are in series so their equivalent resistance i.e. total resistance of the cireceit\begin{array}{c}R_{T}=R_{6}+R_{7}=6.86+5.54 \\R_{T}=12.4 \Omega\end{array}(b) Total current\begin{aligned}I_{T}=\frac{V}{R_{T}} & =\frac{12}{12.4} \\& =0.97 \mathrm{~A}\end{aligned} \mid \begin{array}{l}V=\text { source voltuge } \\=12 \mathrm{~V}\end{array}current through RI_{1}=\frac{R_{2} I_{T}}{R_{1}+R_{2}}=\frac{16 \times 0.97}{12+16}=0.55 \mathrm{~A}Voltage drop across R_{1}V_{1}=I_{1} R_{1}=0.55 \times 12=6.6 \mathrm{~V}\therefore The potential difference across the resistor R_{y}\begin{aligned}V_{y}=V-V_{1} & =12-6.6 \\o, V_{y} & =5.4 \mathrm{~V}\end{aligned}(e) current through R_{2}\begin{aligned}I_{2} & =I_{T}-I_{1} \\& =0.97-0.55 \\\text { or, } I_{2} & =0.42 \mathrm{~A}\end{aligned}(d) Total power dissipated by the circuit\begin{aligned}P_{T} & =I_{T}^{2} R_{T} \\& =(0.97)^{2} \times 12.4 \\\therefore P_{T} & =11.7 \mathrm{~W}\end{aligned}(e) Potential difference across R_{3}v_{3}=v_{4}=5.4 \mathrm{~V}\therefore Power dissipated by R_{3}\begin{aligned}P_{3}=\frac{V_{3}^{2}}{R_{3}}=\frac{(5.4)^{2}}{24} \\\text { or, } P_{3}=1.22 \mathrm{w}\end{aligned}(f) ...