Science Answer the questions.[Phy-5M]

Lets the p.d. across R₁, R₂ and R₃ is V₁, V₂ and V₃ respectively.

s.t. V = V₁ + V₃ + V₂ .........(1)

Let the equivalent resistance be R and current flowing through whole circuit is 1.

By ohm's law

$\frac{\text{V}}{\text{I}}=\text{R}$

V = I × R .....(2)

Applying ohm's law to both R₁, R₂ and R₃.

V₁ = I × R₁ ......(3)

V₂ = I × R₂ ..........(4)

V₃ = I × R₃ ......(5)

From eq. (1), (2), (3), (4) and (5), we get

I × R = I × R₁ + I × R₂ + I × R₃

I × R = I × (R₁ + R₂ + R₃)

R = R₁ + R₂ + R₃

1. For minimum current these two should be connected in series. For maximum current these two should be connected in parallel.

2. In series,

Total resistance = 5 + 10 = 15 ohms

Therefore total current drawn $= \frac{\text{V}}{\text{R}}=\frac{6}{15}=0.4\ \text{amps}$

In parallel,

Total resistance R is given as

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{5}+\frac{1}{10}$

$\frac{1}{\text{R}}=\frac{3}{10}$

$\text{R}=\frac{10}{3}\ \text{ohm}$

Therefore total current drawn by the circuit $=\frac{\text{V}}{\text{R}}=\frac{6}{\big(\frac{10}{3}\big)}=1.8\ \text{amps}.$

1. Total current I = 1 amp is entering the parallel combination of R1 and R. Let I1 current flows through R1 and I2 current flows through R2

Then

$\text{I}_1=\frac{\text{IR}_2}{\text{R}_1+\text{R}_2}$

$=\frac{1\times15}{10+15}=0.6\text{A}$

$\text{I}_2=\frac{\text{IR}_2}{\text{R}_1+\text{R}_2}$

$=\frac{1\times10}{10+15}=0.4\text{A}$

2. p.d. across AB = IR₃ = 1 × 5 = 5V

Equivalent resisyance between B and C is

$\frac{1}{\text{R}'}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}=\frac{1}{10}+\frac{1}{15}$

$\frac{1}{\text{R}'}=\frac{5}{30}$

R' = 6 ohm

Total resistance between A and C is R = 5 + 6 = 11 ohm

p.d. across AC = IR = 1 × 11 = 11V

3. Total resistance = R₃ + R' = 5 + 6 = 11 ohm.

1. $\text{P}.\text{d}.\frac{\text{Work done}}{\text{Charge Moved}}$

Work done = P.d. × Charge moved

= 12 × 1 = 12J

Amount of electrical energy changed into heat and light = 12J

2. Work done = P.d. × Charge moved

= 12 × 5 = 60J

Amount of electrical energy changed into heat and light = 60J

3. $\text{I}=\frac{\text{Q}}{\text{t}}$

Q × I × t

= 2 × 10 = 20C

Work done = p.d. × Charged moved

= 12 × 20 - 240J

Amount of electrical energy changed into heat and light = 240J

1. Total resistance of two resistors that are connected in parallel is:

$\frac{1}{\text{R}'}=\frac{1}{3}+\frac{1}{6}$

$\frac{1}{\text{R}'}=\frac{3}{6}$

R’ = 2ohms

Total resistance of the circuit = 2 + 4 ohms = 6 ohms

2. Total current flowing through the circuit $=\frac{\text{V}}{\text{Total resistance}}$

$\text{I}=\frac{12}{6}=2\ \text{amps}$

3. Potential difference across R₁ = R₁ × I = 4 × 2 = 8V.

Resultant resistance for parallel circuit = R

$\frac{1}{\text{R}}=\frac{1}{6}+\frac{1}{6}$

$\frac{1}{\text{R}}=\frac{2}{6}$

$\text{R} = 3$

Effective resistance = 6 + 3 = 9 ohms

Resultant resistance for each parallel circuit = R

$\frac{1}{\text{R}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$

$\frac{1}{\text{R}}=\frac{3}{6}$

$\text{R} = 2$

Therefore effective resistance = 2 + 2 = 4 ohms.

1. Equivalent resistance of $10\Omega$ and $40\Omega$ resistance (Connected in parallel) is R₁ given as:

$\frac{1}{\text{R}_1}=\frac{1}{10}+\frac{1}{40}=\frac{5}{40}$

$\text{R}_1=8\Omega$

Equivalent resistance of $30\Omega, 20\Omega$ and $60\Omega$ resistance (connected in parallel) is R₂ given as:

$\frac{1}{\text{R}_2}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{6}{60}$

$\text{R}_2=10\Omega$

R₁ and R₂ are connected in series.

$\therefore$ Total resistance in the circuit is R = R₁ + R₂ = 8 + 10 = 18

2. Total current flowing in the circuit, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{12}{18}=0.67\text{A}$

1. Current through R1 $=\text{I}_1=\frac{\text{V}}{\text{R}_1}=\frac{6}{5}=1.2\text{A}$

Current through $\text{R}_2=\text{I}_2=\frac{\text{V}}{\text{R}_2}=\frac{6}{10}=0.6\text{A}$

Current through $\text{R}_3=\text{I}_3=\frac{\text{V}}{\text{R}_3}=\frac{6}{30}=0.2\text{A}$

2. Total current in the circuit = 1.2 + 0.6 + 0.2 = 2A
3. Effective resistance R is given as

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}+\frac{1}{\text{R}_3}$

$=\frac{1}{5}+\frac{1}{10}+\frac{1}{30}$

$=\frac{6+3+1}{30}=\frac{10}{30}$

$\text{R}=\frac{30}{10}=3\Omega$

1. I = ?

Power = VI = 250 × I

4000 = 250I

I = 16amp

2. R = ?

P = I²R

P = 16² × R

$\text{R}=\frac{4000}{16^2}$

R = 15.25 ohm

3. Energy consumed in two hour = P × t

= 4 × 2

= 8Kw - hr

4. If 1KWh = Rs. 4.6

total cost = 8 × 4.6 = Rs. 36.8

1. Series combination

When two or more resistance are connected end to end consecutively, they are said to be connected in series combination. The combined resistance if any number of resistance connected in series in equal to the sum of the individual reisistance.

R = R₁ + R₂.......

2. Parallel combination

When two or more resistance are connected between the same two points, they are said to be connected in parallel combination. The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistance.

$\frac{1}{\text{R}}=\frac{1}{\text{R} _1}+\frac{1}{\text{R}_2}+......$

1. R₁ = 500 ohm, R₂ = 1000 ohm

As per given figure,

R = R₁ + R₂ = 500 + 1000 = 1500 ohm.

2. R₁ = 2 ohm, R₂ = 2 ohm

As per given figure,

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{2}+\frac{1}{2}$

R = 1 ohm

3. R₁ = 4 ohm, R₂ = 4 ohm, R₃

As per given figure,

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}$

R = 2 ohm

Total resistance = R + R₃

= 2 + 3 = 5 ohm

1. Given V = 220V

R_A = R_B = 24 ohm

Current drawn when only coil A is used:

$\text{I}=\frac{\text{V}}{\text{R}_\text{A}}=\frac{220}{24}$

= 9.16 amps

2. Current drawn when coils A and B are used in series:

Total resistance, R = R_A + R_B = 24 + 24 = 48 ohm

$\text{I}=\frac{\text{V}}{\text{R}}=\frac{220}{48}$

= 4.58 amps

3. Current drawn when coils A and B are used in parallel:

Total resistance, $\frac{1}{\text{R}}=\frac{1}{\text{R}_\text{A}}+\frac{1}{\text{R}_\text{B}}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}=\frac{1}{12}$

R = 12 ohm

$\text{I}=\frac{\text{V}}{\text{R}}=\frac{220}{12}$

$=18.33\ \text{amps}$

2. Since the graph is a straight line passing through the origin, so current is directly proportional to the potential difference.

Hence, the ration $\frac{\text{V}}{\text{I}}$ remains constant.

From graph when V = 1.5 volt, I = 0.6 amp

So, $\frac{\text{V}}{\text{I}}=-\frac{15}{0.6}-25\Omega$

For p.d. 0.8V, 1.2V and 1.6V, the value of $\frac{\text{V}}{\text{I}}$ ratio the same i.e., 2.5 ohm.

3. The resistance of the wire is equal to the ratio of potential difference applied and the current passing through it.

$\text{R}-\frac{\text{V}}{\text{I}}-25\Omega$

1. Total resistance of the circuit = R₁ + R₂ = 20 + 4 = 24 ohm
2. We know that

V = IR

Therefore

6 = I × 24

$\text{I}=\frac{6}{24}=0.25\text{amp}$

3. p.d. across bulb = IR₁ = 0.25 × 20 5V
4. p.d. across resistance wire = IR₂ = 0.25 × 4 1V

2. Effective resistance = 20 + 20 = 40 ohms.
3. Current flowing through the circuit $=\text{I}=\frac{\text{V}}{\text{R}}=\frac{5}{40}=0.125\ \text{amps}.$
4. p.d. across each resistance = I × R = 0.125 × 20 = 2.5V