Question F.3: series circuit has two or more resistors end to end. The tot......
series circuit has two or more resistors end to end. The total resistance is equal to the sum of the individual resistances (R_{T}=R_{1}+R_{2}). Also, the sum of the voltage drops across all resistors will equal the total supply voltage (V_{S}=V_{R1}+ V_{R2}).
Find the current in the circuit (I), the voltage across R_{1}\left(V_{R1}\right) and the voltage across R_{2}\left(V_{R2}\right) in Figure F–3.
R_{T}=8\,\mathrm{k}\Omega\,+\,2\,\mathrm{k}\Omega\,=\,10\,\mathrm{k}\Omega
I={\frac{10\,\mathrm{V}}{10\,\mathrm{k\Omega}}}=\mathrm{1\,mA}
V_{R1}\,=\,1\,\mathrm{mA}\,\times\,8\,\mathrm{k \Omega}\,=\,8\,\mathrm{V}
V_{R2}=\,1\,\mathrm{mA}\times2\,\mathrm{k}\Omega\,=\,2\,\mathrm{V}
Check:
V_{S}=8\,\mathrm{V}\,+\,2\,\mathrm{V}\,=\,10\,\mathrm{V}
Notice that the voltage across any resistor in the series circuit is proportional to the size of the resistor. That fact is used in developing the voltagedivider equation:
V_{R2}=V_{S}\times\frac{R_{2}}{R_{1}+R_{2}}
=\;10\;\mathrm{V}\times\frac{2\,\mathrm{k}\Omega}{2\,\mathrm{k}\Omega\,+\,8\,\mathrm{k}\Omega}
= 2 \;V