WBBSE Class 10 Physical Science Chapter 6 Current Electricity Long Answer Questions

Class 10 Physical Science Solution WBBSE Chapter 6 Current Electricity Broad Answer Type Questions

Question 1. State Joule’s laws of heating effect of current.
Answer:

Joule’s Laws (1841):

1.  First Law:

The amount of heat produced in a conductor in a given interval of time of proportion to the square of the current passed. Thus if H be the amount of heat generated in a conductor having resistance R when current 1 passes through it in time t, then

⇒ Hα I² (When R and t are kept constant)

2. Second law:

The amount of heat produced by a given current in a given time is proportional of the resistance of the conductor.

⇒  H α R (When I and t are kept constant)

3. Third Law:

The amount of heat produced in a given conductor by a given current is proportional to the time for which the current passes.

⇒  Hα t (When I and R are kept constant)

“WBBSE Class 10 Physical Science Chapter 6 long answer questions, Current Electricity”

Combining the three laws, we have:

⇒ H α I² RT (When 1, R and t vary) or,

H = \(\frac{I^2 R T}{J}\)

IRT (J= mechanical equivalent of heat = 4.2 joule (calorie)

If I am in ampere, R in ohm, t in second, and H in calorie, then

H= \(\frac{I^2 R T}{J}\)

H= 0.241 RT calorie.

Question 2. Short Note: 

1. Ohmic resistance
2. Non-ohmic resistance

Answer:

1. Ohmic Resistance :

The resistors, which obey Ohm’s Law are said to have ohmic resistances.

Example: All metal or metallic alloy.

2. Non-ohmic resistance :

The resistors which do not obey Ohm’s law are said to have non-ohmic resistance.

Example: Electronic value.

“Class 10 WBBSE Physical Science Chapter 6 long answer questions, Current Electricity study material“

Question 3. Two resistors of 30W and 60W are connected in parallel in an electric circuit. How does the current passing through the two resistors compare?
Answer:

Given

Two resistors of 30W and 60W are connected in parallel in an electric circuit

The potential difference across 30W = B potential difference across 60W

i.e I₁ R₁= I₂ R₂ Or,

⇒ \(\frac{I_1}{I_2}=\frac{R_2}{R_1}\)

= \(\frac{60 \Omega}{30 \Omega}\)

= 2

Question 4. Short Note-Direct Current.
Answer:

Direct Current:

If a resistor connects with the two terminals of an electric cell, then through the resistor a steady current of constant magnitude flows in the same direction (from the positive pole of the cell to the negative pole) and this is called a direct current.

“WBBSE Class 10 Physical Science Chapter 6, Current Electricity long answer solutions”

Question 5. A resistance of 6 Ω is connected with a cell of em. f. 1.5 and negligible internal resistance calculate the current flowing through it.
Answer:

Given

A resistance of 6 Ω is connected with a cell of em. f. 1.5 and negligible internal resistance

V= IR Or, I= \(\frac{V}{R}\)

= \(\frac{1.5}{6}\)

= \(\frac{1}{4}\)

V = 0.25 Amp

The current flowing through it is 0.25 Amp

Question 6. Find the specific resistance of the material of a wire of length 100cm, area of cross-section 0.2 cm,^2, and resistance 2 ohms.

Answer:

R = P× \(\frac{1}{\mathrm{~A}}\)

∴ P= \(\frac{\mathrm{RA}}{\mathrm{1}}\)

= \(\frac{20 \times 0.2}{100}\)

= 40 ×10^-4  Ohm^-cm

So,

R = 20hm

1 = 100 cm

A = 0.2 cm²

P= ?

“WBBSE Class 10 Current Electricity long answer questions, Physical Science Chapter 6”

Question 7. Find effective resistance of the resistors 2 ohm, 4 ohm, 5 ohm connected in

1. Series
2. Parallel.

Answer:

We know, for series combination equivalent resistance

R= r₁+ r₂+r₃

= 2+4+5

= 11 ohm.

We also know, for parallel combination equivalent resistance.

⇒ \(\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\)

= \(\frac{1}{2}+\frac{1}{4}+\frac{1}{5}\)

= \(\frac{19}{20}\)

∴ \(\frac{1}{R}\)= \(\frac{19}{20}\) Or,

= \(\frac{20}{19}\)

R = 1.05 Ohm

Question 8. Short Note-Electric Circuit.
Answer:

Electric Circuit: A continuous conducting path between the terminals of a source of electricity, is called an electric circuit.

Open electric → An electric circuit in which the flow of current stops, because of an open switch is called an open electric circuit. Closed electric

Open Electric Circuit:

Closed Electric Circuit:

Circuit → An electric circuit in which a current flows continuously, because the switch is closed is called a closed electric circuit.

Question 9. There are two copper wires of equal length. The radius of one is twice the other. Find the ratio of their resistances.

Answer:

Given

There are two copper wires of equal length. The radius of one is twice the other.

We know, R= p \(\frac{1}{A}\)

= \(p \frac{1}{\lambda r^2}\)

(r= radius of the wire)

Since, the length and material are the same.

⇒ \(\mathrm{R} \alpha \frac{1}{\mathrm{r}^2}\)

i.e. \(\frac{R_1}{R_2}=\frac{r_2^2}{r_1^2}=\frac{\left(2 r_1\right)^2}{r_1^2}\)

= 4

∴ The thinner wire has a resistance four times the resistance of the thicker wire.

“Class 10 WBBSE Physical Science Chapter 6, Current Electricity detailed answers”

Question 10. Establish the relation between the emf, terminal voltage and internal resistance.
Answer:

The relation between the emf, terminal voltage and internal resistance

Let, a cell of emf E and internal resistance r is connected to an external resistance R

The total resistance of the circuit = R+r

So, the current drawn from the cells

I= \(\frac{e . m . f \text { of the cell }}{\text { total resistance }}\)

I= \(\frac{E}{R+r}\)

E= I (R+r)

The terminal voltage of the cell, V = IR voltage (v) drops due to internal resistance -I As the work is done carrying a unit positive charge once through a complete circuit,

E = V + v

v = E-V

Internal resistance r = \(\frac{V}{I}\)

=  \(\frac{E-V}{I}\)

= \(\frac{E-V}{V / R}\)

= \(R \cdot\left[\frac{E}{V}-1\right]\)

Question 11. A current of 0.5 ampere passes through a wire of resistance 2.5 ohm. for 1 hour. Find the heat produced.
Answer:

Given

A current of 0.5 ampere passes through a wire of resistance 2.5 ohm. for 1 hour.

We know,

H  = \(\frac{I^2 R T}{J}\)

= \(\frac{I^2 R T}{4^2}\)

∴ H \(\frac{(0.5)^2 \times 2.5 \times 3600}{4^2}\)

= 535.7 caloric

So,

I = 0.5 ampere

R = 2.5 ohm

t = 1hr.

= 3600 sec.

H =?

“WBBSE Class 10 Physical Science Chapter 6, Current Electricity long answer key”

Question 12. The resistance of a wire of cross-section area 0.01 cm2 is 10 ohm. What is the length of the wire? The specific resistance of the wire is 50 × 10-6 ohm-cm.
Answer:

Given

The resistance of a wire of cross-section area 0.01 cm2 is 10 ohm.

We know,

R = p = \(\frac{\ell}{\mathrm{A}}\) Or,

l= \(\frac{\mathrm{RA}}{p}\) Or,

l= \(\frac{10 \times 0.01}{50 \times 10^{-6}}\)

∴ l= 2000 cm

So,

R= 10 Ohm

A= 0.01 cm²

p= 50 × 10^-6

Ohm^–cm

1= ?

Question 13. What is the difference between e.m.f and p.d?
Answer:

Difference between e.m.f. and p.d. : e.m.f (electromotive force):

Question 14. A resistance of 6Ω is connected with a cell of e.m.f., 1.5 V, and negligible internal resistance, Calculate the current flowing through it.

Answer:

Given

A resistance of 6Ω is connected with a cell of e.m.f., 1.5 V, and negligible internal resistance

V = IR

I = \(\frac{V}{R}\)

I = \(\frac{1.5}{6}\)

I = \(\frac{1}{4}\)

= 0.25 Amp.

“Class 10 WBBSE Physical Science Chapter 6, Current Electricity important long answer questions”

Question 15. A cell of e.m.f. 1.8v is connected to an external resistance of 22 when p.d. recorded at its terminal is 1.6v. Find the internal resistance of cell.

Answer:

Given

A cell of e.m.f. 1.8v is connected to an external resistance of 22 when p.d. recorded at its terminal is 1.6v.

E = 1.8V, V 1.6 volt, R = 2.2

Internal resistance = \(\frac{R(E-V)}{V}\)

= \( \frac{2(1.8-1.6)}{1.6}\)

= 0.25.Ω