A resistor of 2.4 ohms is connected in series with another of 3.2 ohms. What resistance must be placed across the one of 2.4 ohms so that the total resistance of the circuit shall be 5 ohms?
Answers
Answer:
Resistance = 7.2 ohms
Explanation:
Given:
R1 = 2.4 ohms
R2 = 3.2 ohms
Solution:
2.4 in parallel w/ Rx = 5 - 3.2 = 1.8 ohms, then
1/2.4 + 1/Rx = 1/1.8, from which
1/Rx = 1/1.8 - 1/2.4 = 0.13888
Hence, resistor to be connected in series across the 2.4 ohmm resistor
Rx = 1/0.13888
= 7.2 ohms
Given : A resistor of 2.4 ohms is connected in series with another of 3.2 ohms.
To Find : resistance must be placed across the one of 2.4 ohms so that the total resistance of the circuit shall be 5 ohms
Solution:
Let say Resistance R is placed in parallel to resistor of 2.4 ohms
Hence their equivalent resistance
= 1/(1/R + 1/2.4)
= 2.4R/(R + 2.4)
2.4R/(R + 2.4) is in series with 3.2 ohms
Hence total resistance = 2.4R/(R + 2.4) + 3.2 ohms
total resistance of the circuit shall be 5 ohms
=> 2.4R/(R + 2.4) + 3.2 = 5
=> 2.4R/(R + 2.4) = 1.8
=> 2.4R = 1.8R + 1.8 * 2.4
=> 0.6R = 1.8 * 2.4
=> R = 3 * 2.4
=> R = 7.2 ohms
7.2 ohms resistance must be placed across the one of 2.4 ohms so that the total resistance of the circuit shall be 5 ohms
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