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I have a question about the circuit as shown up. My question is if the total current is 4 Amperes, why do the resistors need a higher current?

Am I missing something? If that so, what it is?

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    \$\begingroup\$ Your calculations for I1 & I2 are not correct. R1 & R2 do not have 64V across them because there is a 48V drop across R3 (4A x 12 ohms = 48V). \$\endgroup\$ – brhans Sep 8 '17 at 2:04
  • \$\begingroup\$ To expand on brhans comment. The voltage across R1 and R2 is only 16 volts (64 times (4 /16)). Then the current through R1 is 16/12, or 1 1/3 amps, and the current through R2 is 16/6, or 2 2/3 amps. Total current is 1 1/3 plus, 2 2/3, or 4 amps. \$\endgroup\$ – WhatRoughBeast Sep 8 '17 at 3:11
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    \$\begingroup\$ What do you mean, the resistors need a higher current? They don't. \$\endgroup\$ – Brian Drummond Sep 8 '17 at 9:30
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You've already calculated the parallel resistance of R1 and R2, put those in series with R3, and calculated the total current correctly.

Now to finish off, you need to use the correct voltage drop for each series resistor. Using Ohm's Law the other way round, the voltage drop is given by \$V=IR\$

For: R3, \$12\Omega\$ * 4A = 48v
For: R1 || R2, \$4\Omega\$ * 4A = 16v

Together, the 16v and 48v add up to 64v, the total input to your circuit.

In an entirely series circuit, one in which the current is constant throughout, the voltage across each component is proportional to the resistance of that component.

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  • \$\begingroup\$ thx to anonymous reviewer who fixed error! \$\endgroup\$ – Neil_UK Sep 8 '17 at 5:48

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