# Class 10 Electricity Important Questions : Solutions

Important Questions Notes**Electricity Important Questions : Solutions**

**The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V.****Sol:**

Here, V = 60 V and I = 4 A

Using, R = V/I = 60/4 = 15 Ω

Now, V = 120 V

So, I = V/R = 120/15 = 8 A**The V-I graph of three resistors is shown below. Which resistor has maximum resistance**?**Sol:**

Resistor R_{3}has maximum resistance.**An electric iron is marked 400 W and 220 V. What is the resistance when iron is hot? How long could it be used for ₹3 of electrical energy that costs 15 paise per unit.****Sol:**

Given, P = 400 W and V = 220 V

Using, P =^{V²}/_{R}

⇒ R =^{V²}/_{P}=^{220²}/_{400}= 121 Ω

So, resistance when iron is hot = 121 Ω

Now, rate of electrical energy = 15 paise per unit

and total money = ₹3 = 300 paise

So, electrical energy consumed, E =^{300}/_{15}= 20 unit = 20 kWh = 20000 Wh

Using, E = P × t

⇒ t =^{E}/_{P}=^{20000}/_{400}= 50 h

So, It can be used for 50 hours.**A 20 cm long uniform wire of resistance 5 Ω is stretched to a uniform wire of 40 cm length. What will be the resistance of new wire?****Sol:**

length = l = 20 cm

Area of cross-section = A

Resistance = R =^{ρl}/_{A}= 5 Ω

New length = l’ = 40 cm = 2l

New area of cross-section = A’

When the wire is stretched, its volume remains the same.

So, A’ × l’ = A × l

⇒ A’ × 40 = A × 20

⇒ A’ =^{A}/_{2}

New resistance

= R’

=^{ρl’}/_{A’}

=^{ρ(2l)}/_{(A/2})

= 4^{ρl}/_{A}

= 4R = 4 × 5 = 20 Ω

Therefore, resistance of the new wire is 20 Ω.**If R**_{1}and R_{2}are the resistances of filaments of a 400 W and a 200 W lamp, designed to operate on the same voltage, then find the relationship between R_{1}and R_{2}.**Sol:**

For R_{1}, P_{1}= 400 W

For R_{2}, P_{2}= 200 W

Using, P =^{V²}/_{R}

(V_{1})^{2}= P_{1}x R_{1}= 400R_{1}

and (V_{2})^{2}= P_{2}x R_{2}= 200R_{2}

Since, both filaments operate on the same voltage.

So, (V_{1})^{2}= (V_{2})^{2}

⇒ 400R_{1}= 200R_{2}

⇒ 2R_{1}= R_{2}**.****State Ohm’s law and write its expression.****Sol:**

According to Ohm’s law, at constant temperature, the potential difference across the ends of a conductor is directly proportional to the amount of current flowing through it.

Mathematically,

V ∝ I**V = IR**where, R is constant of proportionality & is known as resistance.**Draw a circuit diagram to verify Ohm’s law.****Sol:**Circuit diagram for the verification of Ohm’s law:

**The rating of bulb B**_{1}is 60 V, 12 W & bulb B_{2}is 100 V, 100 W. Find the maximum emf of the battery so that all bulbs remain safe.**Sol:**

For B_{1},

V_{1}= 60 V, P_{1}= 120 W

R_{1}=^{V1²}/_{P1}=^{60²}/_{120}= 300 Ω

For B_{2},

V_{2}= 100 V, P_{2}= 100 W

R_{2}=^{V2²}/_{P2}=^{100²}/_{100}= 100 Ω

Since, B_{1}and B_{2}are connected in parallel.

So,^{1}/_{Rp}=^{1}/_{R1}+^{1}/_{R2}=^{1}/_{300}+^{1}/_{100}=^{4}/_{300}

⇒ R_{p }=^{300}/_{4}= 75 Ω

So, R_{eq }across circuit = 75 + 300 + 25 = 400 Ω

Current that can flow through B_{1}=^{V1}/_{R1}=^{60}/_{300}=^{1}/_{5}A

Current that can flow through B_{2}=^{V2}/_{R2}=^{100}/_{100}= 1 A

So, maximum current that can flow through the circuit so that all bulbs remain safe =^{1}/_{5}A

Therefore, required e.m.f. = IR_{eq}=^{1}/_{5}x 400 = 80 V**Find equivalent resistance between P and Q**:**Sol:****Find equivalent resistance between P and Q**:**Sol:**

Here, all resistors are connected in parallel.

So,^{1}/_{Rp}=^{1}/_{R1}+^{1}/_{R2}+^{1}/_{R3}=^{1}/_{10}+^{1}/_{10}+^{1}/_{10}=^{3}/_{10}

⇒ R_{p}=^{10}/_{3}Ω**In reference to the given figure, find the current flowing through 20 Ω resistor**.**Sol:**

R_{AB}= R_{1}+ R_{2}= 5 + 20 = 25 Ω

R_{CD}= R_{1}+ R_{2}= 10 + 40 = 50 Ω^{1}/_{Req}=^{1}/_{RAB}+^{1}/_{RCD}=^{1}/_{25}+^{1}/_{50}=^{3}/_{50}

⇒ R_{eq}=^{50}/_{3}Ω

I = 3 A

So, V = IR_{eq}= 3 x^{50}/_{3}=50 V

In parallel, voltage remains same.

So, current flowing through AB, I =^{V}/_{RAB}=^{50}/_{25}= 2 A

In series, current flowing through each resistor remains same.

So, current flowing through 20 Ω resistor = 2 A.

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