6Ω
9Ω
12Ω
15Ω
Explanation
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Using this formula
V=IR
12 = 2 * R
Make R Subject
R = 12/2 = 6
The total resistance of the circuit is
R(total)= 6
Now resolving the resistance
two resistance are in series while two are in parallel
to make it easier for you
you resolve the one in parallel so it can be in series with the other two
1/R = 1/3 + 1/x
find the LCM of the denominator which is 3 and x = 3x
1/R = x + 3/ 3x
You inverse
R= 3x/ x+3
The Resistance in parallel has been resolved to series, so you can add it up with the other two in series
Now recall that the other two is 2 and 1.5
R(total), 6 = 2 + 1.5 + 3x/ x+3
The LCM of the denominator is x + 3
simplify: 6 = 3.5x + 10.5+ 3x / x+3
cross multiply
6(x+3) = 3.5x + 10.5 + 3x
6x + 18 = 3.5x + 10.5 + 3x
Collect like terms
18 - 10.5 = 3.5x + 3x - 6x
7.5 = 0.5x
Divide each side by the coefficient of x (0.5)
7.5/0.5 = 0.5x/0.5
15 = x
X is 15
