The resistance R3 & R4 are connected in series = (30 + 70) = 100Ω Now three resistor i.e 100Ω, 100Ω & 50Ω is connected in parallel 1/Rp = 1/100 + 1/100 + 1/50 Rp = 100/4 = 25 Ω Resistance 50Ω and 25Ω are connected in series. Rtotal = (50 + 25)Ω Rtotal = 75Ω
Galvanized steel conductors do not corrode, and possess high resistance. Hence such Wires are used in telecommunications circuits, earth wires, guard wire, stray wire, etc.
Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss. At full load copper Loss = I2R At half load copper Loss = (I/2)2 × R = I2/4 × R 400 = I2/4 × R I2R = 4 × 400 I2R = Full load copper Loss = 1600 W
From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below.
Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R
Here
Current I = 1/2 A = 0.5 A
Voltage V = 5 V
R = V/I = 0.5/5
R = 10Ω
EXAMIANS
