# Thread: What is the potential difference across

1. ## What is the potential difference across

Question:

What is the potential difference across (i) AB (ii) BC.

Attempt:
(i) $\displaystyle \frac{200}{400} \times 6 = 3V$

Is this right?

2. Hi

No, it's wrong : the two $\displaystyle 100\Omega$ resistors are in parallel, not in series. Hence $\displaystyle R_{BC}$ is given by $\displaystyle \frac{1}{R_{BC}}=\frac{1}{100}+\frac{1}{100}$.

3. nope i am afraid you are wrong there.

first you need to find the effective resistance of the resistors in parallel(the 100 ohm resistors). Lets say we call them $\displaystyle R_1$
and $\displaystyle R_2$

$\displaystyle \frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$
$\displaystyle \frac{1}{R_{eff}} = \frac{1}{100} + \frac{1}{100}$
$\displaystyle \frac{1}{R_{eff}} = \frac{1}{50}$
$\displaystyle R_{eff} =50 ohm$

Then you can use voltage divider rule to obtain the voltages.

for example voltage across AB
$\displaystyle = \frac{R_{AB}}{R_{Total}} V_{IN}$

$\displaystyle = \frac{200}{200+50} \cdot 6$

$\displaystyle = 4.8V$

try doing for V across BC

4. Originally Posted by looi76
Question:

What is the potential difference across (i) AB (ii) BC.

Attempt:
(i) $\displaystyle \frac{200}{400} \times 6 = 3V$

Is this right?
Here is an extra long explanation on how to do it.

When the resistors are connected in parallel, the current at the splitting node divides. The amount of division depends on how much each branch resists. Since both are 100 ohms, the current will split equally.

Now lets say current 'I' is flowing through the loop, then at the branch each hundred ohm carries 'I/2' current. Since the potential difference is IR, a current 'I/2' through 'R' produces the same potential difference as a current 'I' flowing through a resistance 'R/2'.

Since R here is 100 ohm, R/2 is 50 ohm. This means potential drop across AB is 200I and BC is 50I. The overall drop is therefore 250I. But the overall drop is equal to the applied voltage, which is 6. So 250I = 6 and this means 200I = 4.8V and 50I = 1.2V. This means potential drop across AB is 4.8V and BC is 1.2V.