An electric heater draws 3.5 A from a 110 V source. The resistance of the heatin

Question

An electric heater draws 3.5 A from a 110 V source. The resistance of the heating element is approximately. Three 10 k Ohm resistors are connected in series. A 20 k Ohm resistor is connected in parallel across one of the 10 k Ohm resistors. The voltage source is 24 V. The total current in the circuit is: A 12 k Ohm resistor, a 15 k Ohm resistor, and a 2 k Ohm resistor are in A series with two 10 k Ohm resistors that are in parallel. The source voltage is 75 V. Current through the 15 k Ohm resistor is approximately: Find I_2. 4I_1 + 4I_2 = 2

Explanation / Answer

total resistances of the connection

= (12+15+22) k-ohm in parrale with 10 k-ohm

= 49x10/(49+10) = 8.305 k-ohm

current through the circuit = 75/8.305x10^3 = 9.03x10^-3 A = 9.03mA

THE CURRENT THROUGH 15k-ohm resistance

= 75/49x10^3 = 1.53x10^-3 A = 1.53mA

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