CBSE 10th Standard Science Subject Electricity Chapter Case Study Questions 2021
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CBSE 10th Standard Science Subject Electricity Case Study Questions 2021
10th Standard CBSE
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Reg.No. :
Science
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If two or more resistances are connected in such a way that the same potential difference gets applied to each of them, then they are said to be connected in parallel. The. current flowing through the two resistances in parallel is, however, not the same. When we have two or more resistances joined in parallel to one another, then the same current gets additional paths to flow and the overall resistance decreases. The equivalent resistance is given by \(\begin{equation} \frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} \end{equation}\)
(i) Three resistances, 2 \(\begin{equation} \Omega \end{equation}\) , 6 \(\begin{equation} \Omega \end{equation}\) and 8 \(\begin{equation} \Omega \end{equation}\) are connected in parallel, then the equivalent resistance is
(a) less than 6 \(\begin{equation} \Omega \end{equation}\) but more than 2 \(\begin{equation} \Omega \end{equation}\)
(b) less than 8 \(\begin{equation} \Omega \end{equation}\) but more than 6 \(\begin{equation} \Omega \end{equation}\)
(c) less than 2 \(\begin{equation} \Omega \end{equation}\)
(d) more than 8 \(\begin{equation} \Omega \end{equation}\)
(ii) A wire of resistance 12 \(\begin{equation} \Omega \end{equation}\) is cut into three equal pieces and then twisted their ends together, the equivalent resistance is(a) \(\begin{equation} \frac{3}{8} \Omega \end{equation}\) (b) \(\begin{equation} \frac{4}{3} \Omega \end{equation}\) (c) \(\begin{equation} \frac{3}{4} \Omega \end{equation}\) (d) \(\begin{equation} \frac{5}{6} \Omega \end{equation}\) (iii) Three resistances are connected as shown. The equivalent resistance between A and B is
(a) \(\begin{equation} \frac{2}{3} \Omega \end{equation}\) (b) \(\begin{equation} \frac{3}{2} \Omega \end{equation}\) (c) \(\begin{equation} \frac{4}{3} \Omega \end{equation}\) (d) \(\begin{equation} \frac{3}{4} \Omega \end{equation}\) (iv) Which of the following relation is correct?
(a) I1 = 2I2 = 3I3 (b) I1 = 4I2 = 3I3 (c) 2I1 = I2 = 3I3 (d) 3I1 = 2I2 = I3 (v) Find the current in each resistance.
(a) 1 A (b) 2 A (c) 3 A (d) 0.25 A -
Several resistors may be combined to form a network. The combination should have two end points to connect it with a battery or other circuit elements. When the resistances are connected in series, the current in each resistance is same but the potential difference is different in each resistor. When the resistances are connected in parallel, the voltage drop across each resistance is same but the current is different in each resistor.
(i) The household circuits are connected in(a) series combination (b) parallel combination (c) both (a) and (b) (d) none of these (ii) The two wires of each of resistance R, initially connected in series and then in parallel. In the graph it shows the resistance in series and in parallel. Which of the following is correct?
(a) A denotes parallel combination.
(b) B denotes series combination.
(c) A denotes series combination and B denotes parallel combination.
(d) None of these.
(iii) The equivalent resistance of r1 and r2, when connected in series is R1 and when they are connected in parallel is R2. Then the ratio is(a) \(\begin{equation} \frac{r_{1}}{r_{2}} \end{equation}\) (b) \(\begin{equation} \frac{r_{1}+r_{2}}{r_{1} r_{2}} \end{equation}\) (c) \(\begin{equation} \frac{\left(r_{1}+r_{2}\right)^{2}}{r_{1} r_{2}} \end{equation}\) (d) \(\begin{equation} \frac{r_{1} r_{2}}{2 r_{1}+2 r_{2}} \end{equation}\) (iv) The equivalent resistance between A and B is
(a) 6 \(\begin{equation} \Omega \end{equation}\) (b) 9 \(\begin{equation} \Omega \end{equation}\) (c) 3 \(\begin{equation} \Omega \end{equation}\) (d) 12 \(\begin{equation} \Omega \end{equation}\) (v) Two resistances 10 \(\begin{equation} \Omega \end{equation}\) and 3 \(\begin{equation} \Omega \end{equation}\)are connected in parallel across a battery. If there is a current of 0.2 A in 10 .Q resistor, the voltage supplied by battery is
(a) 2 V (b) 4 V (c) 1 V (d) 8 V -
The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows.
The mathematical expression is given by H = I2Rt.
The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current.
(i) What are the properties of heating element?
(a) High resistance, high melting point
(b) Low resistance, high melting point
(c) Low resistance, high melting point
(d) Low resistance, low melting point.
(ii) What are the properties of electric fuse?
(a) Low resistance, low melting point
(b) High resistance, high melting point.
(c) High resistance, low melting point
(d) Low resistance, high melting point
(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is(a) doubled (b) halved (c) four times (d) one fourth times (iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is
(a) 4 times (b) 2 times (c) \(\begin{equation} \frac{1}{2} \text { times } \end{equation}\) (d) \(\begin{equation} \frac{1}{4} \text { times } \end{equation}\) (v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 \(\begin{equation} \Omega \end{equation}\), the amount of heat produced is
(a) 250 J (b) 5000J (c) 750J (d) 1000J -
The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule. Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved. So for commercial purposes we use a bigger unit of electrical energy which is called kilowatt hour. 1 kilowatt-hour is equal to 3.6 x 106 joules of electrical energy.
(i) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is(a) doubled (b) half (c) remains same (d) four times (ii) The power of a lamp is 60 W The energy consumed in 1 minute is
(a) 360J (b) 36J (c) 3600J (d) 3.6 J (iii) The electrical refrigerator rated 400 W operates 8 hours a day. The cost of electrical energy is \(\begin{equation} ₹ \end{equation}\)5 per kWh. Find the cost of running the refrigerator for one day?
(a) \(\begin{equation} ₹ \end{equation}\)32 (b) \(\begin{equation} ₹ \end{equation}\)16 (c) \(\begin{equation} ₹ \end{equation}\)8 (d) \(\begin{equation} ₹ \end{equation}\)4 (iv) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 \(\begin{equation} \Omega \end{equation}\) for 30 minutes?
(a) 90 kJ (b) 80 kJ (c) 60 kJ (d) 40 kJ (v) Which of the following is correct?
(a) 1 watt hour = 3600 J
(b) lkWh = 36x106J
(c) Energy (in kWh) = power (in W) x time (in hr)
(d) \(\begin{equation} \text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000} \end{equation}\)
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CBSE 10th Standard Science Subject Electricity Case Study Questions 2021 Answer Keys
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(i) (c) :The equivalent resistance in the parallel combination is lesser than the least value, of the individual resistance.
(ii) (b): Resistance of each piece \(\begin{equation} =\frac{12}{3}=4 \Omega \end{equation}\)
\(\begin{equation} \frac{1}{R_{P}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \Rightarrow R_{p}=\frac{4}{3} \Omega \end{equation}\)
(iii) (a): All the three resistors are in parallel.
\(\begin{equation} \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{1}=\frac{1+2+6}{6}=\frac{9}{6} \end{equation}\)
\(\begin{equation} R_{P}=\frac{6}{9}=\frac{2}{3} \Omega \end{equation}\)
(iv) (a): Voltage is same across each resistance.
So, I1 x 5 = I2 X 10 = 15 x I3
I1 = 2I2 = 3I3
(v) (d): All are in parallel.
\(\begin{equation} \begin{array}{l} \frac{1}{R_{p}}=\frac{1}{12} \times 4=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \\ I=\frac{3}{3}=1 \mathrm{~A} \end{array} \end{equation}\)
So, current in each resistor \(\begin{equation} I^{\prime}=\frac{3}{12}=\frac{1}{4} \mathrm{~A} \end{equation}\) -
(i) (b)
(ii) (c): In series combination, resistance is maximum and in parallel combination, resistance is mcm.
(iii) (c) : R1 = r1 + r2
\(\begin{equation} \begin{array}{l} R_{2}=\frac{r_{1} r_{2}}{r_{1}+r_{2}} \\ \frac{R_{1}}{R_{2}}=\frac{\left(r_{1}+r_{2}\right)^{2}}{r_{1} r_{2}} \end{array} \end{equation}\)
(iv) (c): In the given circuit, 3 \(\begin{equation} \Omega \end{equation}\) resistors are in series. RS = 3 + 3 = 6 \(\begin{equation} \Omega \end{equation}\)
Now, RS and 6 \(\begin{equation} \Omega \end{equation}\) are parallel.
\(\begin{equation} \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \end{equation}\)
(v) (a): V = 0.2 x 10= 2 V
So, total voltage supplied is same as 2 V. -
(i) (b)
(ii) (c)
(iii) (a): Given: H = I2Rt
\(\begin{equation} \mathrm{So}, H^{\prime}=(2 I)^{2} \cdot \frac{R}{2} t=2 H \end{equation}\)
(iv) (b): Given: 1= 5 A, resistance = R. Let r be the new radius.
Now,H=I2Rt
Also H' = I'2 R' t
From (i) and (ii), \(\begin{equation} 5^{2} \times \rho \frac{L}{\pi r^{2}} t=10^{2} \times \rho \frac{L}{\pi r^{\prime 2}} \cdot t \end{equation}\)
\(\begin{equation} \frac{25}{r^{2}}=\frac{100}{r^{\prime 2}} \Rightarrow \frac{r^{\prime}}{r}=2 \Rightarrow r^{\prime}=2 r \end{equation}\)
(v) (c): Given: \(\begin{equation} I=0.5 \mathrm{~A}, R=10 \Omega, t=5 \mathrm{~min} \end{equation}\)
\(\begin{equation} \begin{array}{l} H=I^{2} R t=0.5 \times 0.5 \times 10 \times 5 \times 60 \\ H=750 \mathrm{~J} \end{array} \end{equation}\) -
(i) (a): \(\begin{equation} E \propto t \end{equation}\)
(ii) (c): Given: P = 60W, t = 1 min
E = 60 x 1 x 60 = 3600J
(iii) (b): Given: P = 400\(\begin{equation} \Omega \end{equation}\), t = 8 hour
E = 400 x 8 = 3200Wh = 3.2kWh
Cost = 3.2 x 5 = \(\begin{equation} ₹ \end{equation}\) 16
(iv) (a): Given: I= 5 A, R = 2 \(\begin{equation} \Omega \end{equation}\) , t = 30 min
E = I2Rt = 5 x 5 x 2 x 30 x 60
E = 90000J = 90 kJ
(v) (a): 1watt hr = 3600J
