Science And Technology Solutions Solutions for Class 9 Science Chapter 3 Current Electricity are provided here with simple stepbystep explanations. These solutions for Current Electricity are extremely popular among Class 9 students for Science Current Electricity Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science And Technology Solutions Book of Class 9 Science Chapter 3 are provided here for you for free. You will also love the adfree experience on Meritnation’s Science And Technology Solutions Solutions. All Science And Technology Solutions Solutions for class Class 9 Science are prepared by experts and are 100% accurate.
Page No 44:
Question 1:
A. By which method are the appliances connected?
Answer:
A. The appliances are connected in parallel.
B. The potential difference across each appliance should be same as they are connected in parallel. The potential difference between live and neutral wire in domestic wiring is maintained at 220 V in India. Hence, the potential difference across each appliance should be 220 V.
C. No, the current passing through each appliance is not same. From Ohm's law, the current flowing through each appliance is given as
$I=\frac{V}{R}$
Since, potential difference across each appliance is same, thus
$I\propto \frac{1}{R}$
Now, each appliance has different value of resistance. Thus, the current flowing through them will be different. The one with greater value of resistance will receive lesser amount of current and the one with less value of resistance will receive more current.
D. Domestic appliance are connected in parallel because this connection assures that even if one or more appliances get faulty or stops working, the working of others will remain unaffected.
E. The working of other appliances will remain unaffected even if the T.V. stops working. The current through T.V. will stop but the other appliances will receive the approximately the same amount of current as before. This is because in parallel connection, the current divides itself in the number of electrical branches present. If one of the branches becomes open circuited due to some fault, the remaining branches will still be forming complete circuit for the flow of current.
Page No 44:
Question 2:
Which law can you prove with the help of the above circuit?
Answer:
Ohm's law can be proved with the help of above circuit.
Page No 44:
Question 3:
A. Which method should he use to connect the bulbs?
Answer:
When Umesh connects the bulbs one at a time, the filaments of the bulbs get burnt. This means the heat energy generated in the filament is beyond what it can tolerate. Hence, we need to reduce the heat generated across the filament. Now, the heat generated in a resistor is proportional to the amount of current flowing through it. Thus, we need to reduce the current flowing through the bulbs.
A. If bulbs are connected in series, the effective resistance of the circuit increases and hence current through the bulbs decreases. In this way, both the bulbs can be saved from burning.
B. Following are the characteristics of series connection:
 The effective resistance of the circuit is the sum of individual resistance of the resistors present in the circuit.
 The effective resistance of the series circuit is always larger than the greatest resistance in the circuit.
 Current through various electrical components connected in series is same.
 The source voltage gets divided across the electrical components connected in series.
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Question 4:
The following table shows current in Amperes and potential difference in Volts.
a. Find the average resistance.
V (Volts)  I (Amp) 
4  9 
5  11.25 
6  13.5 
Answer:
a. From Ohm's law, we have
$R=\frac{V}{I}$
V (volts) 
I (amp)  R (ohm) 
4  9  $\frac{4}{9}$ 
5  11.25  $\frac{4}{9}$ 
6  13.5  $\frac{4}{9}$ 
b. The nature of the graph between the current and potential difference will be a straight line.
c. The graph proves Ohm's law. Ohm's law states that, under constant temperature and physical conditions, the current flowing through a conductor is always proportional to the potential difference across it i.e.
$V\propto I$
or, V = RI
where R is a constant of proportionality called resistance of the resistor. It tends to resist the flow of charge through a conducting wire. Its SI unit is ohm (Ω).
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Question 5:
‘A’ Group ‘B’ Group
Answer:
‘A’ Group  ‘B’ Group 
1. Free electrons  c. Weakly attached 
2. Current  a. V/ R 
3. Resistivity  d. VA/LI 
4. Resistances in series

b. Increases the resistance in the circuit 
Page No 45:
Question 6:
The resistance of a conductor of length x is r. If its area of crosssection is a, what is its resistivity? What is its unit?
Answer:
Resistivity of a conductor, $\rho =R\frac{A}{L}$
Here, L = x, R = r and A = a
Therefore, $\rho =r\frac{a}{x}$
The unit of resistivity is ohmmetre ($\mathrm{\Omega}\mathrm{m}$).
Page No 45:
Question 7:
Resistances R_{1} , R_{2 }, R_{3} and R_{4} are connected as shown in the figure. S_{1} and S_{2 }are two keys. Discuss the current flowing in the circuit in the following cases.
a. Both S_{1} and S_{2} are closed.
b. Both S_{1} and S_{2} are open.
c. S_{1} is closed but S_{2 }is open.
Answer:
a. The circuit for this case is shown below.
Here, R_{1} and R_{2} are in parallel. Their effective resistance is
$R\text{'}=\frac{{R}_{1}\times {R}_{2}}{{R}_{1}+{R}_{2}}$
Also, R_{4} is in parallel with a conductor of 0 resistance. Their equivalent resistance is therefore 0. The given circuit thus becomes as shown below.
Now, R' and R_{3} are in series. Thus, the equivalent resistance of the circuit is
${R}_{\mathrm{eq}}=\frac{{R}_{1}\times {R}_{2}}{{R}_{1}+{R}_{2}}+{R}_{3}=\frac{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{3}{R}_{2}}{{R}_{1}+{R}_{2}}$
Hence, the current flowing in the circuit is
$I=\frac{V}{\frac{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{3}{R}_{2}}{{R}_{1}+{R}_{2}}}=\frac{V({R}_{1}+{R}_{2})}{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{3}{R}_{2}}$
where, V is the potential difference of the battery.
b. The circuit for this case is shown below.
Here, R_{1}, R_{4} and R_{3} are in series. Therefore, the equivalent resistance of the circuit is
${R}_{\mathrm{eq}}={R}_{1}+{R}_{2}+{R}_{3}$
Hence, the current flowing in the circuit is
$I=\frac{V}{{R}_{\mathrm{eq}}}=\frac{V}{{R}_{1}+{R}_{2}+{R}_{3}}$
where, V is the potential difference of the battery.
c. The circuit for this case is shown below.
Here, R_{1} and R_{2} are in parallel. Their effective resistance is
$R\text{'}=\frac{{R}_{1}\times {R}_{2}}{{R}_{1}+{R}_{2}}$
Now, R_{3}, R' and R_{4} are in series. Thus, the equivalent resistance of the circuit is
${R}_{\mathrm{eq}}=\frac{{R}_{1}\times {R}_{2}}{{R}_{1}+{R}_{2}}+{R}_{3}+{R}_{4}=\frac{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{3}{R}_{2}+{R}_{1}{R}_{4}+{R}_{4}{R}_{2}}{{R}_{1}+{R}_{2}}$
Hence, the current flowing in the circuit is
$I=\frac{V}{\frac{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{3}{R}_{2}+{R}_{1}{R}_{4}+{R}_{4}{R}_{2}}{{R}_{1}+{R}_{2}}}=\frac{V({R}_{1}+{R}_{2})}{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{3}{R}_{2}+{R}_{1}{R}_{4}+{R}_{4}{R}_{2}}$
where, V is the potential difference of the battery.
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Question 8:
f. $\mathrm{x}=\frac{1}{{\displaystyle \frac{1}{{\mathrm{x}}_{1}}}+{\displaystyle \frac{1}{{\mathrm{x}}_{2}}}+{\displaystyle \frac{1}{{\mathrm{x}}_{3}}}}$
Answer:
a. In this case, x_{1} , x_{2 }and x_{3 }are connected in series.
b. In this case, x_{1} , x_{2 }and x_{3 }are connected in series.
c. In this case, x_{1} , x_{2 }and x_{3 }are connected in parallel.
d. In this case, x_{1} , x_{2 }and x_{3 }are connected in parallel.
e. In this case, x_{1} , x_{2 }and x_{3 }are connected in series.
f. In this case, x_{1} , x_{2 }and x_{3 }are connected in parallel.
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Question 9:
A. The resistance of a 1 m long nichrome wire is 6 $\mathrm{\Omega}$. If we reduce the length of the wire to 70 cm. what will its resistance be?
B. When two resistors are connected in series, their effective resistance is 80 $\mathrm{\Omega}$. When they are connected in parallel, their effective resistance is 20 $\mathrm{\Omega}$. What are the values of the two resistances?
C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current?
Answer:
A. Resistance of a wire is directly proportional to its length. Let R_{1 }and R_{2} be the resistances of the nichrome wire at 100 cm (1 m) and 70 cm, respectively. Therefore,
${R}_{1}\propto 100.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{R}_{2}\propto 70.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\left(\mathrm{ii}\right)\mathrm{by}\hspace{0.17em}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{R}_{2}}{{R}_{1}}=\frac{70}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{\mathit{2}}=\frac{70}{100}\times {R}_{1}$
Now, R_{1 }= 6 $\mathrm{\Omega}$
$\Rightarrow {R}_{\mathit{2}}=\frac{70}{100}\times 6=4.2\mathrm{\Omega}$
B. Let the two resistances be R_{1} and R_{2}. Therefore,
R_{1} + R_{2 }= 80 $\mathrm{\Omega}$
or, R_{1 }= 80 ${R}_{2}$ .....(i)
and
$\frac{1}{{R}_{1}}+\frac{{\displaystyle 1}}{{\displaystyle {R}_{2}}}=\frac{{\displaystyle 1}}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{1}{R}_{2}=20({R}_{1}+{R}_{2})\phantom{\rule{0ex}{0ex}}\mathrm{As},{R}_{1}=80{R}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow (80{R}_{2}){R}_{2}=20(80{R}_{2}+{R}_{2})\phantom{\rule{0ex}{0ex}}80{R}_{2}^{2}{R}_{2}=1600\phantom{\rule{0ex}{0ex}}{R}_{2}^{2}80{R}_{2}+1600=0\phantom{\rule{0ex}{0ex}}({R}_{2}40)({R}_{2}40)=0\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{2}=40\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{R}_{1}=8040=40\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$
c. Current flowing through the conducting wire is
$I=\frac{\mathrm{Charge}\left(Q\right)}{\mathrm{Time}\left(t\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{420}{5\times 60}=1.4\mathrm{A}$
Hence, the current flowing through the wire is 1.4 A.
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