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Science · 2022 · 3 marks
CBSE 2022 · Region 4 · Set 1 · Q13
An electric motor rated $\displaystyle 1100$ W is connected to $\displaystyle 220$ V mains. Find :(i)The current drawn from the mains,(ii)Electric energy consumed if the motor is used for $\displaystyle 5$ hours daily for $\displaystyle 6$ days.(iii)Total cost of energy consumed if the rate of one unit is ₹ $\displaystyle 5$ .OR Study the following circuit and find :
(i)Effective resistance of the circuit(ii)Current drawn from the battery(iii)Potential difference across the $\displaystyle 5 \Omega$ resistor
An electric motor rated $\displaystyle 1100$ W is connected to $\displaystyle 220$ V mains. Find :
(i)
The current drawn from the mains,
(ii)
Electric energy consumed if the motor is used for $\displaystyle 5$ hours daily for $\displaystyle 6$ days.
(iii)
Total cost of energy consumed if the rate of one unit is ₹ $\displaystyle 5$ .
OR Study the following circuit and find :
(i)
Effective resistance of the circuit
(ii)
Current drawn from the battery
(iii)
Potential difference across the $\displaystyle 5 \Omega$ resistor
Marking-scheme solution
Power $\displaystyle (\mathrm{P})=1100 \mathrm{~W}, \mathrm{~V}=220 \mathrm{~V}$
(i)
\[\begin{aligned}
\text { Current drawn } & =I=\frac{P}{V} \\
& =\frac{1100 \mathrm{~W}}{220 \mathrm{~V}}=5 \mathrm{~A}
\end{aligned}
\]
(ii)
\[E=P \times t
\]
\[=1100 \mathrm{~W} \times 5 \mathrm{~h} \times 6=33000 \mathrm{~Wh}
\]
Cost of one commercial unit $\displaystyle =₹ 5$
Energy consumed $\displaystyle =33 \mathrm{kWh}=33$ unit $\displaystyle =118.8 \times 10^{6} \mathrm{~J}$
Cost of $\displaystyle 33$ unit $\displaystyle =33 \times 5=₹ 165$
Effective resistance of the circuit
\[\begin{array}{l}
R_{s}=R_{3}+R_{4}=4 \Omega+6 \Omega=10 \Omega \\
\frac{1}{R_{p}}=\frac{1}{R_{s}}+\frac{1}{R_{2}}=\frac{1}{10 \Omega}+\frac{1}{10 \Omega}=\frac{2}{10 \Omega}=\frac{1}{5 \Omega} \\
R_{p}=5 \Omega
\end{array}
\]
Total resistance of the circuit $\displaystyle =R_{1}+R_{p}+R_{5}=5+5+10=20 \Omega$
(ii)
Current drawn from the battery
\[\begin{array}{l}
V=20 \mathrm{~V}, R=20 \Omega \\
I=\frac{V}{R}=\frac{20 \mathrm{~V}}{20 \Omega} \\
I=1 \mathrm{~A}
\end{array}
\]
(iii)
Reading in voltmeter connected across $\displaystyle 5 \Omega$ Resistance
\[\begin{array}{l}
V=I R \\
\mathrm{I}=1 \mathrm{~A} \\
\mathrm{R}=5 \Omega \\
\mathrm{~V}=1 \mathrm{~A} \times 5 \Omega=5 \mathrm{~V}
\end{array}
\]
SECTION-C
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