Given:
the voltage of the battery is
[tex]V=9\text{ V}[/tex]
six resistance are connected in the circuit.
Explanation:
resistor a and b are in series connection. so they can be added in the following way
[tex]\begin{gathered} R=R_6+R_2 \\ R=6+2 \\ R_{ab}=8\text{ }\Omega \end{gathered}[/tex]
now resistor Rab ohm and Re ohms are in parallel combination.
they can be added in the following way
[tex]\begin{gathered} R=\frac{R_8\times R_4}{R_8+R_4} \\ R=\frac{8\times4}{12} \\ R_{abe}=2.66\text{ }\Omega \end{gathered}[/tex]
now all the resistor are in series combination. they can be added as
[tex]\begin{gathered} R_{equi}=6+2.66+1+3 \\ R_{equi}=12.66\text{ }\Omega \end{gathered}[/tex]
above is the equivalent resistance of the circuit.
now calculate the current in the circuit.
apply ohm's law
[tex]V=IR_{equi}[/tex]
Plugging all the values in the above relation, we get
[tex]\begin{gathered} I=\frac{9\text{ V}}{12.66\text{ }\Omega} \\ I=0.71\text{ A} \end{gathered}[/tex]
above is the total current flowing in the circuit.
now calculate the voltage and current across each resistor.
voltage across the resistor d is
[tex]\begin{gathered} V=IR_d \\ V=0.71\text{ A}\times3\text{ }\Omega \\ V=2.13\text{ V} \end{gathered}[/tex]
now voltage across resistor f is
[tex]\begin{gathered} V=IR_f \\ V=0.71\times6 \\ V=4.26\text{ V} \end{gathered}[/tex]
this amount of voltage will be same across the resistor ab and resistor e.
voltage across the resistor e is
[tex]V=4.26\text{ V}[/tex]
voltage across the resistor c is
[tex]\begin{gathered} V=0.71\times1 \\ V=0.71\text{ V} \end{gathered}[/tex]
now calculate the current across Rab and Re.
the current in across Rab is
[tex]\begin{gathered} V=I_{ab}\times R_{ab} \\ I_{ab}=\frac{1.88}{8} \\ I_{ab}=0.24\text{ A} \end{gathered}[/tex]
now the current across Re is
[tex]\begin{gathered} V=I_e\times R_e \\ I_e=\frac{V}{R_e} \\ I_e=\frac{1.88}{4} \\ I_e=0.47\text{ A} \end{gathered}[/tex]
