# Analysing circuits question

Question asks to work out the potential drop across each resistor and the current going through each resistor.

For R1 I got a current of 0.5A and a voltage drop of 3V. For R2 I got a current of 0.3A and a voltage drop of 6V. For R3 I got a current of 0.2A and a voltage drop of 6V.

Just not sure if I have done it properly, could someone please let me know if these answers are wrong.

• Those figures are correct. – Steve G May 4 '16 at 12:05
• Are you sure that the internal resistance of the battery doesn't matter? – Daniel Tork May 4 '16 at 12:50

Just practicing my $\LaTeX$ stuff....

R2 || R3 = $\dfrac{1}{\frac{1}{20}+ \frac{1}{30}}$ = 12 ohm.

Hence total resistance seen by 9V supply is 12 ohm + 6 ohm = 18 ohm hence current = 0.5 A.

The rest is just ohms law.

Using the parallel resistor formula for $R_2$ and $R_3$ which is $R_{23}=\frac{R_2R_3}{R_2+R_3}$ ,we obtain $R_{23}$=12 ohms.To get the equivalent resistance:$R_{ec}$=$R_1$+$R_{23}$(since $R_1$ and $R_{23}$ are in series).Next,the current is obtained by $I=\frac{U}{R_{ec}}$,so I=0,5 A.The voltage drop is now calculated easily:$U_1$=$I$$R_1$=3V and $U_{23}$=$I$$R_{23}$=6V.
Now if you split the $R_{23}$ resistor back into the resistors that form it,you will know that the voltage drop is the same across $R_2$ and $R_3$ and is equal to $U_{23}$,so $U_2$=$U_3$=$U_{23}$=6V.You can find out what is the current through each of the 2 resistors using the U=IR formula.So $I_2=\frac{U_2}{R_2}$=0,3A and $I_3=\frac{U_3}{R_3}$=0,2A.
Make sure the exercise doesn't want you to make use of the internal voltage drop of the battery.Also,if you want to make sure you solved exercises like this correctly for the future,you can check if all/some of the equations are true(say $I=I_2+I_3$ so 0,5=0,5 which is true).You've got it right.You can solve this using Kirchhoff's laws,too.