In the circuit shown in the figure, find the magnitude of current in the upper branch. [Solved]

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 10(3.33/8.33) = 4v
total voltage across 10 is 6 volts. current is 6/10 = 0.6 amp (c)
Now that means voltage across the upper 5 ohms is 106 = 4 volts, so current is 5/4 = 1.25 amp. (a)
Voltage across middle 5 ohms is 56 = 1v, so current is 1/5 = 0.2 amps, pointing to left (b)
The rest is easy. calculate the voltages across the 3 and 4 ohm resistors, and then it is just addition

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 5(3.33/8.33) = 2v

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 5(3.33/8.33) = 2v

first combine the series resistors, 5 in the top and 5 in the bottom.
Use superposition

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 5(3.33/8.33) = 2v

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 10(3.33/8.33) = 4v
total voltage across 10 is 6 volts. current is 6/10 = 0.6 amp (c)
Now that means voltage across the upper 5 ohms is 106 = 4 volts, so current is 5/4 = 1.25 amp. (a)
Voltage across middle 5 ohms is 56 = 1v, so current is 1/5 = 0.2 amps, pointing to left (b)
The rest is easy. calculate the voltages across the 3 and 4 ohm resistors, and then it is just addition

Get Answer for  For the reaction kclo2kcl o2 kclo2kcl o2 assign oxidation numbers to each element on each side of the equation. k in kclo2: [Solved]

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 5(3.33/8.33) = 2v

now there is 5 and 10 in parallel, total 50/15 = 3.33. That is in series with 5. Voltage across the 10 by voltage divider is 10(3.33/8.33) = 4v
total voltage across 10 is 6 volts. current is 6/10 = 0.6 amp (c)
Now that means voltage across the upper 5 ohms is 106 = 4 volts, so current is 5/4 = 1.25 amp. (a)
Voltage across middle 5 ohms is 56 = 1v, so current is 1/5 = 0.2 amps, pointing to left (b)
The rest is easy. calculate the voltages across the 3 and 4 ohm resistors, and then it is just addition

This post is last updated on hrtanswers.com at Date : 1st of September – 2022