Question

For the following circuit, find the total resistance and current, then find the currentand voltage across each resistor.

Answer 1

Given:

the voltage of the battery is

`V=9\text{ V}`

six resistance are connected in the circuit.

Step-by-step explanation:

resistor a and b are in series connection. so they can be added in the following way

`\begin{gathered} R=R_6+R_2 \\ R=6+2 \\ R_(ab)=8\text{ }\Omega \end{gathered}`

now resistor Rab ohm and Re ohms are in parallel combination.

they can be added in the following way

`\begin{gathered} R=(R_8* R_4)/(R_8+R_4) \\ R=(8*4)/(12) \\ R_(abe)=2.66\text{ }\Omega \end{gathered}`

now all the resistor are in series combination. they can be added as

`\begin{gathered} R_(equi)=6+2.66+1+3 \\ R_(equi)=12.66\text{ }\Omega \end{gathered}`

above is the equivalent resistance of the circuit.

now calculate the current in the circuit.

apply ohm's law

`V=IR_(equi)`

Plugging all the values in the above relation, we get

`\begin{gathered} I=\frac{9\text{ V}}{12.66\text{ }\Omega} \\ I=0.71\text{ A} \end{gathered}`

above is the total current flowing in the circuit.

now calculate the voltage and current across each resistor.

voltage across the resistor d is

`\begin{gathered} V=IR_d \\ V=0.71\text{ A}*3\text{ }\Omega \\ V=2.13\text{ V} \end{gathered}`

now voltage across resistor f is

`\begin{gathered} V=IR_f \\ V=0.71*6 \\ V=4.26\text{ V} \end{gathered}`

this amount of voltage will be same across the resistor ab and resistor e.

voltage across the resistor e is

`V=4.26\text{ V}`

voltage across the resistor c is

`\begin{gathered} V=0.71*1 \\ V=0.71\text{ V} \end{gathered}`

now calculate the current across Rab and Re.

the current in across Rab is

`\begin{gathered} V=I_(ab)* R_(ab) \\ I_(ab)=(1.88)/(8) \\ I_(ab)=0.24\text{ A} \end{gathered}`

now the current across Re is

`\begin{gathered} V=I_e* R_e \\ I_e=(V)/(R_e) \\ I_e=(1.88)/(4) \\ I_e=0.47\text{ A} \end{gathered}`