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Answer: Electric potential is the work done per unit charge to bring a positive test charge from infinity to a point in an electric field without acceleration. Its formula is V = W/q, where V is potential in volt (V), W is work (J), and q is charge (C). You can imagine it like an electric height: a point at higher potential can give more energy to charges, just like a higher hill gives more gravitational energy. More work needed means higher potential. For example, if 20 J is needed to bring 4 C, then V = 20/4 = 5 V. This means every 1 C placed at that point possesses 5 J of electric potential energy.
Answer: Potential difference (P.D.) is the work done per unit charge to move a charge from one point to another. It is the “push” or electric pressure that makes electrons move. Without a difference in potential, no current flows, just as water does not flow if both sides are at the same level. In a cell of 1.5 V, every 1 C of charge gains 1.5 J when moving from the negative to positive terminal. In homes (India), the 220 V difference between the live and neutral wires enables appliances to run by supplying energy to charges. Devices like bulbs, fans, chargers operate only because of this potential difference.
Answer: A voltmeter measures potential difference between two points, so it must be connected in parallel across the component. In parallel, the two terminals of the voltmeter are at the exact points whose voltage we want to measure. A voltmeter has very high resistance so that only a tiny current flows through it, preventing disturbance of the circuit. If you connect it in series, its high resistance greatly reduces the circuit current, the device may not work properly, and the reading becomes meaningless. In some cases, it can even overstress the meter or the circuit. Therefore, remember: voltmeter → parallel connection, high resistance, minimal circuit disturbance.
Answer: Think of potential as water level in a tank. A point at higher level has more pressure to push water, similar to higher electric potential. Water flows only when there is a difference in level; likewise, charges flow only when there is a potential difference. A greater level difference gives faster water flow; a higher potential difference gives more current. In the hill analogy, a ball rolls downhill from higher to lower height; similarly, current flows from higher to lower potential externally. When you increase the voltage across a bulb (say from 1.5 V to 3 V), each coulomb carries more energy, so the bulb converts more energy per second, making it brighter.
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Answer: In a parallel connection, each bulb gets the same potential difference (1.5 V). The difference in brightness comes from their resistances (R). The power converted by each bulb is P = V²/R. If one bulb is brighter, it must have a lower resistance, so for the same V, it converts more energy per second into light and heat. The current through each bulb is I = V/R; the lower-R bulb draws more current, further increasing its power. A voltmeter across each bulb will read 1.5 V, confirming equal potential difference. Thus, brightness in parallel circuits mainly depends on resistance, not on unequal voltages.
Answer: The dryer is designed for 110 V, so internal parts are rated for certain currents and power at that voltage. If connected to 220 V, the potential difference doubles. In many resistive devices, power P = V²/R. Doubling V leads to power becoming roughly four times the designed value. This causes excessive current, rapid heating, possible insulation failure, and risk of smoke, fire, or component burn-out. Motors inside may spin beyond rated speed, worsening mechanical stress. Because the circuit is not built to withstand 220 V, it may fail immediately. Safe operation requires a step-down transformer or a dual-voltage device. Hence, higher potential difference seriously overloads the appliance.
Answer: The reading of 1.5 V across the cell means each 1 C gains 1.5 J of energy from the cell. The 1.5 V across the bulb shows each 1 C loses the same 1.5 J in the bulb as light and heat. This equality illustrates energy conservation: the total energy supplied by the source equals the total energy used by circuit elements in a closed loop. In simple circuits with one cell and one bulb, the entire potential rise of the cell equals the entire potential drop across the bulb. A voltmeter confirms this by measuring the potential difference across each element. Any mismatch would suggest internal resistance or connection errors.
Answer: The mistake is connecting the voltmeter in series. A voltmeter has very high resistance to minimize current through it. In series, this large resistance chokes the circuit current, so the bulb barely glows or does not glow at all. The reading becomes unreliable because the circuit’s behavior has been disturbed. A voltmeter must be connected in parallel across the two terminals of the device whose potential difference we want to measure. This way, it measures the voltage across the bulb without significantly altering the current in the main path. In short: voltmeter → parallel, high resistance; ammeter → series, low resistance.
Answer: When the switch is open, the circuit is broken; both terminals of the open switch can be at the same potential because no current flows and there is no potential drop across ideal wires. Hence, the voltmeter reads 0 V across the switch. When the switch is closed, the circuit is complete; the cell provides a potential rise, and the bulb creates a potential drop equal to the supply (ignoring internal resistance). Therefore, measuring across the bulb now shows the full supply voltage. This demonstrates that potential difference appears across components that resist current (like bulbs), while ideal conductors and open gaps can show 0 V across them in certain configurations due to equal potentials.
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