Physics · Current Electricity · NEET
Adding a parallel resistor gives current a new, extra path. More paths means it is easier for charge to flow, so total resistance drops. Think of doors in a room: adding another door lets more people out per second. That is why R (parallel) is always less than the smallest resistor in the group. If this ever comes out bigger, you made an arithmetic slip.
Voltage is the SAME. Both ends of every resistor connect to the same two junctions, so the potential difference V across each is identical. The CURRENT is what splits (I = I1 + I2 + ...). This is the opposite of series, where current is same and voltage divides. Fixing this one point clears most parallel-circuit errors.
Same V across each resistor. By Ohm's law the branch currents are I1 = V/R1 and I2 = V/R2. By Kirchhoff's junction rule the total current is I = I1 + I2 = V/R1 + V/R2. But the whole combination also obeys I = V/R, where R is the effective resistance. Equate: V/R = V/R1 + V/R2. Cancel V (same everywhere) to get 1/R = 1/R1 + 1/R2.
Only for exactly TWO resistors in parallel: R = (R1·R2)/(R1+R2). For three or more you must add reciprocals: 1/R = 1/R1 + 1/R2 + 1/R3. A common exam trap is using product/sum on three resistors, which is wrong.
They combine to R/n. Example: five 10 ohm resistors in parallel give 10/5 = 2 ohm. This appears again and again in NEET (cut-wire and heater problems), so memorise it.
The effective resistance of four identical wires in parallel is 0.25 ohm. Their effective resistance in series is:
10 resistors (R each) are first connected in series with a battery (emf E, negligible internal resistance); then the same 10 are connected in parallel. The current increases n times. The value of n is:
A wire of resistance R is cut into 8 equal pieces. Two sets are made by joining four pieces in parallel each; the two sets are then connected in series. The net resistance is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
1/R = 1/R1 + 1/R2 + 1/R3 + ... where R is the effective resistance. For just two resistors you can use the shortcut R = (R1 x R2)/(R1 + R2).
Yes. Adding parallel paths always makes it easier for current to flow, so the equivalent resistance is smaller than every individual resistor in the group. If your answer is larger, recheck the reciprocals.
Voltage is the same across every branch. Current divides between the branches, with more current through the smaller resistance.
They give R/n. For example, four 8 ohm resistors in parallel give 8/4 = 2 ohm. This is a high-frequency NEET result.
They test whether you can reduce a network step by step and keep track of which rule (same V or same I) applies. Cut-wire, heater-rating and current-ratio questions appear almost every year in Current Electricity.
