Step 1: Understanding the Concept:
A simple household circuit consists of a power source, a switch to control the flow of current, and a load (the bulb) that consumes electrical energy. The components are connected in series, meaning the current flows through each component one after the other. The switch, when open, breaks the circuit, stopping the current flow and turning the bulb off. When closed, it completes the circuit, allowing current to flow and the bulb to light up.
Step 2: Apparatus Required:
- An electric bulb with a holder
- A single-pole switch
- A power source (like a battery, battery eliminator, or a safe low-voltage AC source)
- Connecting wires
Step 3: Circuit Diagram and Procedure:
Circuit Diagram:
Procedure:
1. Safety First: Ensure the power source is turned off or disconnected before making any connections.
2. Connect the Source to Switch: Take one connecting wire and connect it from the positive terminal of the power source to one of the terminals on the switch.
3. Connect Switch to Bulb: Take another wire and connect it from the second terminal of the switch to one of the terminals on the bulb holder.
4. Complete the Circuit: Take a third wire and connect it from the other terminal of the bulb holder back to the negative terminal of the power source. This completes the series circuit.
5. Testing: Turn on the power source. The bulb should be off. Now, close the switch (flick it to the 'ON' position). The circuit is now complete, and the bulb should light up. Open the switch (flick to 'OFF'), and the bulb should turn off.
Step 4: Result:
A simple household circuit was successfully assembled. The switch correctly controls the operation of the bulb, demonstrating the basic principle of a series electrical circuit.
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.