Advertisements
Advertisements
प्रश्न
The resistance of two resistors joined in series is 8Ω and in parallel is 1.5Ω. Find the value of the two resistances.
Advertisements
उत्तर
In series, R1 + R2 = 8 Ω
In parallel `("R"_1"R"_2)/("R"_1 + "R"_2) = 1.5` Ω
∴ R1R2 = 8 × 1.5 = 12 Ω
Now (R1 - R2)2 = (R1 + R2)2 - 4R1R2
∴ (R1 - R2)2 = (8)2 - 4 × 12
or (R1 - R2)2 = 64 - 48 = 16
or R1 - R2 = 4 Ω
On solving equations (i) and (ii) R1 = 6 Ω and R2 = 2 Ω
संबंधित प्रश्न
Find the expression for the resistors connected in series and write the two characteristics of it. (Draw figure).
Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.
How does the resistance of a pure metal change if its temperature decreases?
How does the resistance of a wire change when:
its diameter is tripled?
Calculate the combined resistance in each case:
State the order of resistivity of (i) a metal, (ii) a semiconductor and (iii) an insulator.
State and explain the laws of resistance.
In the circuit shown below, calculate the equivalent resistance between the points (i) A and B, (ii) C and D.
Two resistors of 4Ω and 6Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
Draw a labelled circuit diagram showing the above arrangement.
