Fundamentals of Electric Circuits and Networks

Solution-9

• Power dissipated in a resistor \[\begin{aligned} &P_{\text {avg }}=V_{\text {rms }}^2 / R \\ &P_{\text {avg }}=\frac{1}{T} \int P d t \\ &P=\frac{1}{10} \frac{1}{\pi}\left[\int_{0}^{2 \pi / 3} 10^{2} d t+\int_{2 \pi / 3}^{\pi}-10^{2} d t\right] \\ &P=\frac{100}{10} \times \frac{1}{\pi}\left[\frac{2 \pi}{3}+\frac{\pi}{3}\right] \\ &\quad=10 \mathrm{~W} \end{aligned}\]

Problem-10

• A wire-wound resistor is to be made from a 0.2 mm diameter constantan wire wound around a cylinder of 1 cm diameter. How many turns are needed for a resistance of 50 \(\Omega\). (Take resistivity of constantan as \(49 \times 10^{-8}~\Omega m\) )

Solution-10

• We know \(\mathrm{R}=\rho \mathrm{L} / \mathrm{A}\)

• Given, \[\begin{aligned} \rho&=49 \times 10^{-8} \Omega \mathrm{m}, \mathrm{R}=50 \Omega, \mathrm{d}=0.2 \mathrm{~mm} \\ \mathrm{~A}&=\pi \times \mathrm{d}^{2} / 4 \\ \quad&=\pi \times(0.2)^{2} / 4 \\ \quad&=0.0314 \mathrm{~mm}^{2}=3.14 \times 10^{-8} \mathrm{~m}^{2} \end{aligned}\]

• So, \(50=49 \times 10^{-8} \times \mathrm{L} /\left(3.14 \times 10^{-8}\right)\) \[\mathrm{L}=3.204 \mathrm{~m}\]

• As the wire is wound around a cylinder of radius \(0.5 \mathrm{~cm}\) \[\begin{aligned} \mathrm{L} &=2 \pi \mathrm{r} \times \mathrm{N} \\ 3.204 &=2 \pi \times 0.005 \times \mathrm{N} \\ \mathrm{N} &=101.98 \approx \mathbf{1 0 2} \text { turns } \end{aligned}\]

Problem-11

• The average value of the current waveform shown in figure

Solution-11

\(\mathrm{i}(\mathrm{t})\) is periodic a period in \(\mathrm{t}\) of \(\pi \mathrm{sec}\). \[\mathrm{i}(\mathrm{t})=\left\{\begin{array}{l} \left(\frac{10 \mathrm{t}}{\pi}+5\right) \quad 0 \leq \mathrm{t} \leq \frac{\pi}{2} \\ \left(-\frac{10 \mathrm{t}}{\pi}+15\right) \frac{\pi}{2} \leq \mathrm{t} \leq \pi \end{array}\right.\]

\[\begin{aligned} I_{\text {ave }} &=\frac{1}{\pi} \int_{t=0}^{t=\frac{\pi}{2}}\left(\frac{10 t}{\pi}+5\right) d t+\int_{t=\frac{\pi}{2}}^{t=\pi}\left(-\frac{10 t}{\pi}+15\right) d t \\ &=\frac{1}{\pi}\left[\frac{5 t^{2}}{\pi}+56\right]_{t=0}^{t=\frac{\pi}{2}}+\left[\frac{-5 t^{2}}{\pi}+15 t\right]_{t=\frac{\pi}{2}}^{t=\pi} \\ &=7.5 \mathrm{~A} \end{aligned}\]

Problem-12

• The domestic power load in a house comprises the following:

  • 8 lamps of 100 W each,

  • 3 fans of 80 W each,

  • 1 refrigerator of 1/2 hp,

  • 1 heater of 1000 W.

  1. Calculate the total current taken from the supply of 230 V

  2. Calculate the energy consumed in a day, if on an average only a quarter of the above load persists all the time.

Solution-12

• The total load is given as

• Current taken from the supply, \(I=\dfrac{P}{V}=\dfrac{2413}{230} \approx 10.5 \mathrm{~A}\)

• Energy consumed per day \(=2413 \mathrm{~W} \times(1 / 4) \times 24=14478 \mathrm{Wh}=\mathbf{1 4 . 4 7 8} \mathbf{~ k W h}\)

Problem-13

• Figure shows the waveform of the current passing through an inductor of resistance 1 \(\Omega\) and inductance 2 H. The energy absorbed by the inductor in the first four seconds is

Solution-13

• Energy dissipated by the Inductance (2H), \[\begin{aligned} &=\left.\frac{1}{2} L i^{2}(t)\right|_{t=0} ^{t=4} \\ &=\frac{1}{2} \times 2 \times\left[i^{2}(4)-i^{2}(0)\right]=36 \text { jouls } \end{aligned}\]

• Energy absorbed by the resistance, \[\begin{aligned} &=\int_{0}^{4} R i^{2}(t) d t =\int_{0}^{2} 9 t^{2} d t+(36 \times 2) \\ &=\left.3 t^{3}\right|_{0} ^{2}+72 \\ &=24+72 \\ &=96 \mathrm{~J} \end{aligned}\]

• Total energy absorbed by the physical inductor \[\begin{aligned} &=36+96 \\ &=132 \mathrm{~J} \end{aligned}\]

Problem-14

• The current entering the positive terminal of a device is \(i(t)=3 e^{-2 t} \mathrm{~A}\) and the voltage across the device is \(v(t)=5 d i / d t \mathrm{~V}\)

  1. Find the charge delivered to the device between \(\mathrm{t}=0\) and \(\mathrm{t}=2 \mathrm{~s}\)

  2. Calculate the power absorbed.

  3. Determine the energy absorbed in \(3 \mathrm{~s}\).

Solution-14

• (a) \[\begin{aligned} \mathrm{q}=\int \mathrm{idt}=\int_{0}^{2} 3 \mathrm{e}^{-2 \mathrm{t}} \mathrm{dt}=\left.\frac{-3}{2} \mathrm{e}^{2 \mathrm{t}}\right|_{0} ^{2} \\ &=-1.5\left(\mathrm{e}^{-4}-1\right)=\\ \underline{1.4725 \mathrm{C}} \end{aligned}\]

• (b) \[\begin{aligned} &\mathrm{v}=\frac{5 \mathrm{di}}{\mathrm{dt}}=-6 \mathrm{e}^{2 \mathrm{t}}(5)=-30 \mathrm{e}^{-2 \mathrm{t}} \\ &\mathrm{p}=\mathrm{vi}=-90 \mathrm{e}^{-4 \mathrm{t}} \mathrm{W} \end{aligned}\]

• (c) \(w=\int \mathrm{pdt}=-90 \int_{0}^{3} \mathrm{e}^{-4 \mathrm{t}} \mathrm{dt}=\left.\frac{-90}{-4} \mathrm{e}^{-4 t}\right|_{0} ^{3}=\underline{-22.5 \mathrm{~J}}\)

Problem-15

• The charge entering the positive terminal of an element is \[q=10 \sin 4 \pi t \mathrm{mC}\] while the voltage across the element (plus to minus) is \[v=2 \cos 4 \pi t \mathrm{~V}\]

  1. Find the power delivered to the element at \(\mathrm{t}=0.3 \mathrm{~s}\)

  2. Calculate the energy delivered to the element between 0 and \(0.6 \mathrm{~s}\).

Solution-15

• Current \[i=\frac{d q}{d t}=40 \pi \cos 4 \pi t \mathrm{~mA}\]

• Power \[\begin{aligned} &p=v i=80 \pi \cos ^{2} 4 \pi t \mathrm{~mW} \\ &\text { At } \mathrm{t}=0.3 \mathrm{~s} \\ & p=80 \pi \cos ^{2}(4 \pi \times 0.3)=\underline{164.5 \mathrm{~mW}} \end{aligned}\]

• Energy \[\begin{aligned} \text { (b) } W=\int p d t=80 \pi \int_{0}^{0.6} \cos ^{2} 4 \pi t d t=40 \pi \int_{0}^{0.6}[1+\cos 8 \pi t] d t \mathrm{~mJ} \\ W=40 \pi\left[0.6+\frac{1}{8 \pi} \sin 8 \pi t \mid \begin{array}{c} 0.6 \\ 0 \end{array}\right]=\underline{78.34 \mathrm{~mJ}} \end{aligned}\]