Solved Problem on Resistors

Between the terminals, A and B of the figure, it is applied a voltage of 200 V. Calculate the magnitudes of the currents in each resistor and the equivalent resistance.

Problem Data:

Potential difference between A and B: V_AB = 220 V.

Solution

The two 10 Ω resistors, placed between points D and B, are connected in parallel (Figure 1). The equivalent resistance for equal resistors is given by

\[ \bbox[#99CCFF,10px] { R_{eq}=\frac{R}{n}} \]

For n = 2

\[ \begin{gather} R_{1}=\frac{10}{2}\\ R_{1}=5\;\Omega \tag{I} \end{gather} \]

Figure 1

Note: We could also determine the equivalent resistance by applying the expression to two resistors in parallel

\[ \bbox[#99CCFF,10px] {R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}} \]

\[ \begin{gather} R_{1}=\frac{10\times 10}{10+10}\\ R_{1}=\frac{10}{20}\\ R_{1}=5\;\Omega \end{gather} \]

Or we could determine the equivalent resistance by applying the general expression for the equivalent resistance in parallel

\[ \bbox[#99CCFF,10px] {\frac{1}{R_{eq}}=\sum_{i=1}^{n}{\frac{1}{R_{i}}}} \]

\[ \begin{gather} \frac{1}{R_{1}}=\frac{1}{10}+\frac{1}{10}\\ \frac{1}{R_{1}}=\frac{10+10}{10\times 10}\\ \frac{1}{R_{1}}=\frac{20}{100}\\ R_{1}=\frac{100}{20}\\ R_{1}=5\;\Omega \end{gather} \]

The two 5 Ω resistors between points C and B are connected in series (Figure 2). The equivalent resistance for equal resistors is given by

\[ \bbox[#99CCFF,10px] {R_{eq}=nR} \]

for n = 2

\[ \begin{gather} R_{2}=2\times 5\\ R_{2}=10\;\Omega \tag{II} \end{gather} \]

Figure 2

Note: We could also determine the equivalent resistance by applying the overall expression for the association of resistors in series

\[ \bbox[#99CCFF,10px] {R_{eq}=\sum_{i=1}^{n}R_{i}} \]

\[ \begin{gather} R_{3}=5+5\\ R_{3}=10\;\Omega \end{gather} \]

For the two parallel resistors between points C and B (Figure 3), we have the same result (I) above

\[ R_{4}=5\;\Omega \]

Figure 3

For the two series resistors between points A and B (Figure 4), we have the same result (II) above, which is the equivalent resistance of the circuit

\[ \bbox[#FFCCCC,10px] {R_{eq}=10\;\Omega} \]

Figure 4

The Ohm's Law is given by

\[ \bbox[#99CCFF,10px] {U=ri} \]

substituting U = V_AB = 220 V e r = R_eq = 10 Ω (Figura 5)

\[ \begin{gather} V_{AB}=R_{eq}i\\ i=\frac{V_{AB}}{R_{eq}}\\ i=\frac{200}{10}\\ i=20\;\text{A} \end{gather} \]

Figure 5

Returning to the circuit of Figure 4, we have two series resistors, in this case, the two resistors are flowed by the same current (Figure 6)

\[ \bbox[#FFCCCC,10px] {i_{1}=20\;\text{A}} \]

\[ i_{2}=20\;\text{A} \]

Figure 6

Returning to the circuit of Figure 3 we have two resistors of the same value connected in parallel between points C and B, in this case, the current i₂ is also divided between the two resistors (Figure 7)

\[ \begin{gather} i_{3}=i_{4}=\frac{i_{2}}{2}\\ i_{3}=i_{4}=\frac{20}{2}\\ i_{3}=i_{4}=10\;\text{A} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {i_{3}=10\;\text{A}} \]

Figure 7

Returning to the circuit of Figure 2, we have two series resistors, in this case, the two resistors are flowed by the same current (Figure 8)

\[ \bbox[#FFCCCC,10px] {i_{5}=10\;\text{A}} \]

\[ i_{6}=10\;\text{A} \]

Figure 8

Returning to the circuit of Figure 1, we have two resistors of the same value connected in parallel between points D and B, in this case, the current i₆ is also divided between the two resistors (Figure 9)

\[ \begin{gather} i_{7}=i_{8}=\frac{i_{6}}{2}\\ i_{7}=i_{8}=\frac{10}{2} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {i_{7}=i_{8}=5\;\text{A}} \]

Figure 9

The complete diagram for the currents will be (Figure 10)

Figure 10

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