# Different ways to calculate resistances

Now, one perspective is that point X is similar to point Y in terms of p.ds, therefore $$R_{BY} = R_{BX} = R/2$$. And this is actually what the answer book says.

However, I thought of it this way, if $$V_{AB}=P$$, we have the same p.d across ZY. Since the combined resistance between B and Y is $$R/2$$, then p.d across BY is $$P/2$$. In addition, if the current through AB and ZY is I, that through CX will be its half. Ultimately, calculating $$R_{BX}$$ gives $${P/2}\over{I/2}$$, i.e $$R$$. Why is this line of thought wrong?

Also, is there a way of calculating this using Kirchhoff's laws?

• The mistake is when you assume p.d. across BY as $P/2$. The current is not the same in BY as AB (or ZY) so you cannot assume that potential will be halved. Take the current to be half too (it symmetrically divides, you yourself have written correctly), and then p.d. across BY is I/2$\times$R. But now you don't know I without knowing the net resistance! Oops. So your 1st method is necessary, using the formula for parallel resistances. Always follow this order: Current depends on net resistance, and voltage across each component is I$\times$R. Commented Feb 17, 2017 at 14:05
• @KalpakGupta when I said BY I meant the effective resistance of the two resistors in parallel. The current throught them collectively is still I, and the resistance is R/2; therefore the p.d is P/2. Commented Feb 17, 2017 at 14:33
• Then while calculating R (BX), you should divide by I, which is the combined current, as R (BX) is the combined resistance. But you're using I/2 which is not the total current, so it gives resistance R of any one resistor. Moreover, you have assumed net resistance to be R/2 while calculating the potential, and then again you use that potential to calculate resistance, so the answer must be same. Be careful, it will turn our consistent. Doubts? Commented Feb 20, 2017 at 5:06

Let $V_{ab}=25$ and $R=10$. You would get $I=1$ because the total resistance is $25 (10+5+10)$. $5$ is because the $2$ resistors are in parallel. Thus the voltage drops around the circuit are $10,5,$and $10$ volts and each resistor in the parallel portion has $0.5$ amps.

• Please use mathjax feature to format equations. Commented Feb 17, 2017 at 16:28

If there were $3$ resistors of equal resistance connected in series to a battery of emf $V$ volts then the potential difference across each resistor would be $\frac{V}{3}$.

In your case, you have $3$ resistors ($2$ of resistance $R$ and $1$ resistor of equivalent resistance $\frac{R}{2}$). So the potential drop across each resistor won't be the same.

$Req = \frac{5R}{2} ;\space I = \frac{2V}{5R}$

The potential drop across AB, BY and YZ will be $\frac{2V}{5}$, $\frac{V}{5}$ and $\frac{2V}{5}$ respectively.

Let the potential drop across AB be $P$ which according to the calculation is equal to $\frac{2V}{5}$.

Then the potential drop across CX or BY will be $\frac{V}{5}$ or $\frac{P}{2}$. As far as the results are concerned, your results are right.

It's correct to assume that in a circuit if 3 resistors ($R$, $R$ and $\frac{R}{2}$) connected in series and current through them is same then, according to ohm's law $V=I.Req$, potential drop across resistor of resistance $R$ will be $\frac{2V}{5}$ (or $I.R$ or $\frac{VR}{Req}$) and $\frac{R}{2}$ will be $\frac{V}{5}$ (or $I.\frac{R}{2}$ or $\frac{VR}{2Req}$).

• Please use mathjax feature to format equations. Commented Feb 17, 2017 at 16:28
• I was looking for the tutorial...Thank you for the help.. Commented Feb 17, 2017 at 16:45