NSEJS #3

The potential drop across the $9ohm$ resistor is $4.5V$ . So, by Ohm's law, current through it (from left to right):
$i = \frac{4.5}{9} = 0.5A$ This is the same current that flows through the $3ohm$ resistor hence the same current flows through their series combination of $9 + 3 = 12ohms$ . Let's establish that current $I$ when comes across two resistors of $12ohms$ and $6ohms$ divides such that the $12ohms$ resistor receives a current of $0.5A$ .

Now, if the equivalent resistance of the $10ohms$ and $15ohms$ resistors is calculated it comes out to be:
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{15} => R_{eq} = 6ohms$

This again creates the situation in which current $I$ has to divide between resistors of resistances $6ohms$ and $12ohms$ . Therefore, the current through the $12ohms$ resistor is $\boxed{0.5A}$ .

let Re = equivalent resistance V = 4.5V R = 9 Ohms I = ?

I = V / R I = 4.5 / 9 I = 0.5A

Re = 1 / (1 / 10 + 1 / 15) Re = 1 / (0.1 + 0.0666) Re = 1 / 0.1666 Re = 6 Ohms

6 + 12 = 18 Ohms 4.5/18 = 0.25A

Current through the 12 Ohm Resistor is 2(0.25), or 0.5 amperes.

As no two resistors has the same value, I’ll name those by theirs values.

• Current in R9; 4,5V / 9Ω = 0,5A
• As current is the same in R3 and R9 so voltage at R3; 3Ω x 0,5A = 1,5V
• Voltage at R(3-9); 4,5V + 1,5V = 6V
• Voltage at R6 is the same as R(9-3) so current in R6; 6V / 6Ω = 1A
• Total current in R(9-3-6); 0,5A + 1A = 1,5A
• Equivalent resistance of R(10-12-15); 1/(1/10Ω + 1/12Ω + 1/15Ω) = 4Ω
• As current in (10-12-15) is the same as in R(9-3-6); 4Ω x 1,5A = 6V
• So current in R12 = 6V / 12Ω = 0,5A QED