Current Electricity – Class 9 Science Question & Answers (Maharashtra Board)

Answer All household appliances are wired together using a parallel connection.

Answer The potential difference across every single appliance is the exact same and matches the main voltage supplied to the house.

Answer No, the current flowing through them is different. Because the voltage is constant but every appliance has its own unique electrical resistance, the current varies depending on that resistance ($I = V/R$).

Answer They are connected in parallel so that every device gets full operating power and can be turned on or off completely independently without affecting the rest of the house.

Answer No, the other appliances will keep working normally because each device has its own independent electrical path.

Which law can you prove with the help of the above circuit?

• Answer

  • Circuit Setup Explanation: In this experimental setup, a resistor and a variable resistor (rheostat) are connected in series with a cell and a plug key. An ammeter is placed in series to measure the current, and a voltmeter is connected in parallel across the resistor to measure the potential difference.

  • Law Proved: This circuit is used to prove Ohm's Law.

Answer He should connect the bulbs in a series combination.

Answer

1. The exact same electric current flows through both bulbs.

2. The total effective resistance of the circuit increases, which keeps the current at a safe level and protects the delicate filaments from burning out.

3. The total voltage is shared between the bulbs.

Answer

• In a series connection, we simply add the two resistances together:

$$\text{Total~Resistance} = 15\ \Omega + 30\ \Omega = \mathbf{45}\ \mathbf{\Omega}$$

• a. Find the average resistance.

  • Simple calculation guide: Resistance is calculated by dividing Voltage by Current ($R = V \div I$).

    • Row 1: $4 \div 9 = 0.44\ \Omega$

    • Row 2: $5 \div 11.25 = 0.44\ \Omega$

    • Row 3: $6 \div 13.5 = 0.44\ \Omega$

  • Taking the average of all three gives $0.44\ \Omega$.

• b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)

Answer The graph will be a straight line passing straight through the origin $(0,0)$.

Answer It proves Ohm's Law. The law states that under constant physical conditions, the current flowing through a conductor is directly proportional to the potential difference across its ends.

• Answer

  • Formula: Resistance is given by $r = \rho\frac{x}{a}$(where $\rho$ is resistivity).

  • Finding Resistivity ($\rho$): Rearranging the formula gives:

$$\rho = \frac{r \cdot a}{x}$$

• Unit: The SI unit of resistivity is ohm-meter ($\Omega \cdot \text{m}$).

Answer Current flows smoothly through the entire circuit, splitting across the parallel branch of $R_{1}$ and $R_{2}$, and then passing through $R_{3}$ and $R_{4}$.

Answer Because both keys are open, the circuit path is broken. No current flows at all.

Answer Opening key $S_{2}$ breaks the main return path back to the power source, meaning no current flows through the circuit, even though $S_{1}$ is closed.

• a. Current I flows through $x_{1},x_{2}$and $x_{3} \rightarrow$Series connection

• b. x is larger than $x_{1},x_{2}$and $x_{3} \rightarrow$Series connection

• c. x is smaller than $x_{1},x_{2}$and $x_{3} \rightarrow$Parallel connection

• d. The potential difference across $x_{1},x_{2}$and $x_{3}$is the same $\rightarrow$Parallel connection

• e. $x = x_{1} + x_{2} + x_{3} \rightarrow$Series connection

• f. $x = \frac{1}{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}}} \rightarrow$Parallel connection

• Beginner-Friendly Logic: Think of resistance like a hallway: a longer hallway creates more resistance to walking, while a shorter hallway is easier. If you shorten the wire, the resistance drops by that exact fraction.

• Step-by-Step Solution:

  1. Note down what is given:

     • Original length ($L_{1}$) = $1\text{~meter} = 100\text{~cm}$(we convert meters to centimeters so units match)

     • Original resistance ($R_{1}$) = $6\ \Omega$

     • New length ($L_{2}$) = $70\text{~cm}$

  2. Set up the proportion formula:

$$\frac{\text{New~Resistance}}{\text{Old~Resistance}} = \frac{\text{New~Length}}{\text{Old~Length}}$$

$$\frac{R_{2}}{6} = \frac{70}{100}$$

3. Calculate:

$$R_{2} = \frac{70}{100} \times 6 = 0.7 \times 6 = \mathbf{4.2}\ \mathbf{\Omega}$$

• Final Answer: The new resistance is $4.2\ \Omega$.

• Beginner-Friendly Logic: We have two hidden numbers. When added, they equal $80$. When combined in parallel, they equal $20$. Let's find them using plain logic.

• Step-by-Step Solution:

  1. Series equation:

$$R_{1} + R_{2} = 80$$

2. Parallel equation:

$$\frac{R_{1} \times R_{2}}{R_{1} + R_{2}} = 20$$

3. Substitute the series total into the parallel equation:

   • Replace $\left( R_{1} + R_{2} \right)$ at the bottom with $80$:

$$\frac{R_{1} \times R_{2}}{80} = 20$$

• Multiply $20$ by $80$ to find their product:

$$R_{1} \times R_{2} = 1600$$

4. Find the individual numbers:

   • What two equal numbers add up to $80$ and multiply to give $1600$?

   • $40 + 40 = 80$, and $40 \times 40 = 1600$.

• Final Answer: The values of the two resistors are $40\ \Omega$and $40\ \Omega$.

• Beginner-Friendly Logic: Current tells us how much charge flows every single second. We just need to divide total charge by total seconds.

• Step-by-Step Solution:

  1. Note down what is given:

     • Charge ($Q$) = $420\text{~Coulombs~(C)}$

     • Time ($t$) = $5\text{~minutes}$

  2. Convert minutes into seconds:

     • Since 1 minute = 60 seconds:

$$5 \times 60 = 300\text{~seconds}$$

3. Apply the current formula ($I = \frac{\text{Charge}}{\text{Time}}$):

$$I = \frac{420}{300}$$

4. Simplify:

   • $\frac{420}{300} = \frac{42}{30} = \mathbf{1.4}\text{~A}$

• Final Answer: The value of the electric current is $1.4\text{~A}$.