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Electricity: Circuits and their Components - A Simple Electrical Circuit

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A simple electric circuit consists of a source of energy (cell), a conducting path (wires), a switch, and an electrical load (bulb). Current flows from the positive terminal to the negative terminal of the cell when the circuit is closed.

A basic circuit diagram showing a cell, a switch, and a bulb connected by wires in a continuous loop.
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The switch acts as a control element; in the 'ON' position, it bridges the gap to make the circuit 'closed' or complete. In the 'OFF' position, it creates a break, resulting in an 'open' circuit where no current flows.

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Electric cells can be combined to form a battery. To increase the total voltage, the positive terminal of one cell must be connected to the negative terminal of the next cell in a series arrangement.

Two cells connected in series where the positive of the first meets the negative of the second.
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A circuit diagram uses standard symbols to represent components. For example, a long thin line represents the positive terminal and a short thick line represents the negative terminal of a cell.

The symbol for an electric cell with polarity markings.

📐Formulae

Vtotal=V1+V2+V3+⋯+VnV_{total} = V_1 + V_2 + V_3 + \dots + V_n

Current Flow=Positive (+)→Negative (-)\text{Current Flow} = \text{Positive (+)} \rightarrow \text{Negative (-)}

💡Examples

Problem 1:

A student wants to make a battery using four cells, each having a voltage of 1.5 V1.5\text{ V}. What will be the total voltage of the battery if they are connected correctly in series?

Solution:

1.51.51.5+1.56.0\begin{array}{r} 1.5 \\ 1.5 \\ 1.5 \\ + 1.5 \\ \hline 6.0 \end{array} The total voltage is 6.0 V6.0\text{ V}.

Explanation:

In a series circuit, the total voltage (VtotalV_{total}) is the sum of the individual voltages of the cells connected. Since there are four cells of 1.5 V1.5\text{ V} each, we add them: 1.5+1.5+1.5+1.5=6.0 V1.5 + 1.5 + 1.5 + 1.5 = 6.0\text{ V}.

Problem 2:

In a circuit with a bulb, a switch, and a battery, the bulb does not glow even when the switch is in the 'ON' position. What could be the possible reasons?

Solution:

Possible reasons:

  1. The bulb might be fused (the filament is broken).
  2. The cells are not connected properly (e.g., ++ to ++ instead of ++ to −-).
  3. The battery is exhausted (no chemical energy left).
  4. There is a loose connection in the wires.

Explanation:

For current to flow and the bulb to glow, the circuit must be closed and all components must be functional and connected in the correct polarity.

Problem 3:

A circuit contains three identical cells, each of 2 V2\text{ V}, connected in series as shown in the diagram. Calculate the total voltage and determine the direction of the current.

Three cells in series connected by a wire loop with an arrow indicating clockwise current flow.

Solution:

Total Voltage Vtotal=2 V+2 V+2 V=6 VV_{total} = 2\text{ V} + 2\text{ V} + 2\text{ V} = 6\text{ V}. The current flows clockwise from the positive terminal to the negative terminal.

Explanation:

When cells are connected in series (positive to negative), their voltages add up. The conventional current always travels from the high potential (positive) to the low potential (negative).

Problem 4:

A simple circuit is constructed using a 9 V9\text{ V} battery, a switch in the 'ON' position, and two identical bulbs connected in series. If one bulb requires a minimum of 5 V5\text{ V} to glow brightly, will the bulbs glow at full brightness? Provide the total voltage and the voltage shared by each bulb.

A circuit diagram showing a 9V battery connected in series with two bulbs and a closed switch.

Solution:

Total Voltage (Vtotal)=9 V\text{Total Voltage } (V_{total}) = 9\text{ V} Voltage per bulb=Vtotal2=9 V2=4.5 V\text{Voltage per bulb} = \frac{V_{total}}{2} = \frac{9\text{ V}}{2} = 4.5\text{ V} Since 4.5 V<5 V4.5\text{ V} < 5\text{ V}, the bulbs will not glow at full brightness.

Explanation:

In a series circuit, the total voltage provided by the battery is divided equally among identical components. Here, the 9 V9\text{ V} is split into two 4.5 V4.5\text{ V} shares. Because each bulb receives less than its required 5 V5\text{ V}, they will appear dim.