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
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Calculate the resistance of an aluminium cable of length 10 km and diameter 2.0 mm if the resistivity of aluminium is 27 × 10−8 Ωm.
You are supplied with a number of 100 Ω resistors. How could you combine some of these resistors to make a 250 Ω resistor?
State how are the two resistors joined with a battery when equivalent resistance is more than either of the two resistances.
Name the following substances:
(i) Showing low resistivity,
(ii) Showing very high resistivity,
(iii) Showing moderate resistivity.
What do you mean by power rating of an electrical appliance? How do you use it to calculate
(a) the resistance of the appliance and
(b) the safe limit of the current in it, while in use?
State expression for Cells connected in parallel.
What is a resistance? Define it with respect to Ohm’s law,
Two electric bulbs have resistances in the ratio 1 : 2. If they are joined in series, the energy consumed in them is in the ratio:
