Tag: electric current and its effects

Questions Related to electric current and its effects

If voltage across a bulb rated $220V-100W$ drops by $2.5$% of its rated value, the percentage of the rated value by which the power would decrease is

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $5$%

  2. $10$%

  3. $20$%

  4. $2.5$%

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Correct Option: A

A padcular ohmmeter uses a battery to provide a potential difference across an unknown resistance  whose value S to be measured. The meter measures the resulting current through this resistor and is calibrated to read out corresponding value of resistance. Suppose that this ohmmeter is used to measure he resistance of a typical incandescent tungsten-filament light bulb. The value of the resistance of the light bulb will be

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. less then when the bulb will be in use in a 120 volt circuit

  2. more then when the bulb will be in use in a 120 volt circuit

  3. the same as then when the bulb will be in use in a 120 volt circuit

  4. more information when needed to determine whether it's A,B and C

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Correct Option: B

A $500\ W$ heating unit is designed to operate on a $115\ V$ line. If line voltage drops to $110\ V$ line, the percentage drop in heat output will be:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $7.6\ \%$

  2. $8.5\ \%$

  3. $8.1\ \%$

  4. $10.2\ \%$

Show answer
Correct Option: B
Explanation:

Given:
$H _1 = 500\ W$
$V _1 = 115\ V$
$V _2 = 110\ V$

From Joule's Law of heating,
$H _1 = \cfrac{(V _1)^2}{R}$ and $H _2 = \cfrac{(V _2)^2}{R}$
$\Rightarrow R = \cfrac{(V _1)^2}{H _1} = \cfrac{(V _2)^2}{H _2}$
$\Rightarrow H _2 = \cfrac{V _2^2}{V _1^2} H _1$
$\therefore H _2 = \cfrac{(110)^2}{(115)^2} (500)$
$\therefore H _2 = 457.46 W$

The percentage drop in heat output will be:
$\cfrac{H _2-H _1}{H _1} \times 100 = \cfrac{500 - 457.46}{500} \times 100 = \cfrac{42.54}{500} \times 100 = 8.5\ \%$

An electric heater operating at $220\ V$ boils $5\ l$ of water in $5\ \text{minutes}$. If it is used on a $110\ V$ line, it will boil the same amount of water in:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $10\ \text{minutes}$

  2. $20\ \text{minutes}$

  3. $5\ \text{minutes}$

  4. $1\ \text{minute}$

Show answer
Correct Option: B
Explanation:
Heat required to boil the water will be same in the two cases.
Heat produced by heater is:
$Q = \cfrac{V^2}{R} t$
$V^2 = \cfrac{QR}{t}$
$V^2 \propto \dfrac{1}{t}$

$\cfrac{t _2}{t _1} = \cfrac{V _1^2}{V _2^2}$
$t _2 = \dfrac{220^2 \times 5}{110^2}$
$t _2 = 20\ min$

The maximum current $I$, which can be passed through a fuse without melting varies with its radius $r$ as:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $I \propto r$

  2. $I \propto r^{3/2}$

  3. $I \propto r^2$

  4. $I \propto (1/r^2)$

Show answer
Correct Option: B
Explanation:

Heat lost per second per unit surface area of fuse wire is
$H = \frac{I^2 \rho }{2 \pi ^2 r^3}$
$\Rightarrow  I^2 \propto r^3$
$\Rightarrow  I \propto r^{\frac{3}{2}}$

If current is flowing through a 10 $\Omega$ resistor, then indicate in which case the maximum heat will be generated?

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. Current of $5\ A$ flows for $2\ \text{minutes}$.

  2. Current of $4\ A$ flows for $3\ \text{minutes}$.

  3. Current of $3\ A$ flows for $6\ \text{minutes}$.

  4. Current of $2\ A$ flows for $5\ \text{minutes}$.

Show answer
Correct Option: C
Explanation:

Heat produced is given by
$Q = I^2 R t$

Option A:
$5\ A$ flows for $2\ \text{minutes}$
$Q = 5^2 \times 10 \times 2 = 500\ J$

Option B:

$4\ A$ flows for $3\ \text{minutes}$

$Q = 4^2 \times 10 \times 3 = 480\ J$


Option C:
$3\ A$ flows for $6\ \text{minutes}$
$Q = 3^2 \times 10 \times 6 = 540\ J$

Option D:
$2\ A$ flows for $5\ \text{minutes}$
 $Q = 2^2 \times 10 \times 5 = 200\ J$


Hence, option C is correct.

The energy expended in $1\  kW$ electric heater in $30\ \text{seconds}$ will be:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $\displaystyle 3\times 10^4\ J$

  2. $\displaystyle 3\times 10^4\ erg$

  3. $\displaystyle 3\times 10^4\ eV$

  4. $0$

Show answer
Correct Option: A
Explanation:

$\text{Energy = Power} \times \text{time}$
$\text{Energy} = 1000 \times 30 = 30000\ J = 3 \times {10}^{4}\ J$

A resistor has resistance R. When the potential difference across the resistor is V, the current in
the resistor is I. The power dissipated in the resistor is P. Work W is done when charge Q flows
through the resistor.
What is not a valid relationship between these variables? 

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $I =\frac {P}{V}$

  2. $Q =\frac {W}{V}$

  3. $R =\frac {P}{I^2}$

  4. $R =\frac {V}{P}$

Show answer
Correct Option: D
Explanation:

We know that 

$P= VI$
$\implies P= V\times \dfrac{V}{R}$
$\implies P = \dfrac{V^2}{R}$
$\implies R= \dfrac{V^2}{P}$..............(1)
Therefore the option D is wrong .

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. The energy supplied in kWh to the three heaters in 5 hours is :

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $3.75 kWh$

  2. $4 kWh$

  3. $0.6 kWh$

  4. $5.74 kWh$

Show answer
Correct Option: A
Explanation:

Power is given as: $P=VI$


Substituting, $I=\dfrac{V}{R}$ in the above formula, we get, $P=\dfrac { { V }^{ 2 } }{ R } $

Given that the voltage is $250\ V$ and the power is $60\ W$, the resistance of the bulb is calculated as follows.

$R=\dfrac { { V }^{ 2 } }{ P } =\dfrac { { 100 }^{ 2 } }{ 250 } =40\Omega$

Hence, the resistance of each resistor is 40 ohms.

The energy consumed by the appliance in kWh is given by the formula: $P=\dfrac { { V }^{ 2 } }{ R } \times t=\dfrac { { 100 }^{ 2 } }{ 40 } \times t= 1.25\ kWh$.

When three heater are connected in parallel, then the total energy consumed is given as $1.25\ kWh\times 3=3.75\ kWh.$.

Hence, the total power consumed is 3.75 kWh.

A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. The cost of energy consumed at Rs. 4.20 per kWh for 5 hours will be :

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $Rs. 250$

  2. $Rs. 300$

  3. $Rs. 310$

  4. $Rs. 315$

Show answer
Correct Option: D
Explanation:

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. 


The power is given by the product of applied voltage and the electric current.

 That is, $P=VI. The\ power\ of\ the\ geyser\ is\ given\ as\ 1500 W$

The energy consumed by the geyser in kWh is given by the formula $Q=P\times t\quad =\quad 1500W\times 50hours\quad =\quad 75\quad kWh$. 

Hence, the energy consumed by the geyser in 5 hours is 75 kWh. 

The cost of energy consumed per kWh is given as Rs. 4.20.

That is, $4.20\times 75\quad kWh=\quad Rs.315$

Hence, the total cost is given as Rs. 315.