What is the amount and the date of renewal of the license?

March 17, 2026Physics

ScanSolve AI Answer

March 17, 2026

Answer

R_1 + R_2 $$

Step 1: Total resistance $R$ .

$R = R_1 + R_2$

$R = 3\,\Omega + 5\,\Omega$

$R = 8\,\Omega$

Step 2: Total current $I$ .

$I = \frac{U}{R}$

$I = \frac{12\,V}{8\,\Omega}$

$I = 1.5\,A$

Step 3: Voltage across $R_1$ , $U_1$ .

$U_1 = I R_1$

$U_1 = 1.5\,A \times 3\,\Omega$

$U_1 = 4.5\,V$

Step 4: Voltage across $R_2$ , $U_2$ .

$U_2 = I R_2$

$U_2 = 1.5\,A \times 5\,\Omega$

$U_2 = 7.5\,V$

Step 5: Power dissipated in $R_1$ , $P_1$ .

$P_1 = U_1 I$

$P_1 = 4.5\,V \times 1.5\,A$

$P_1 = 6.75\,W$

Step 6: Power dissipated in $R_2$ , $P_2$ .

$P_2 = U_2 I$

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Quick Answer

Total resistance R. R = R_1 + R_2 R = 3\, + 5\, R = 8\, Step 2: Total current I.

Asked on March 17, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on March 17, 2026

Step 1: Total resistance R. R = R_1 + R_2 R = 3\, + 5\, R = 8\, Step 2: Total current I. I = (U)/(R) I = 12\,V8\, I = 1.5\,A Step 3: Voltage across R_1, U_1. U_1 = I R_1 U_1 = 1.5\,A × 3\, U_1 = 4.5\,V Step 4: Voltage across R_2, U_2. U_2 = I R_2 U_2 = 1.5\,A × 5\, U_2 = 7.5\,V Step 5: Power dissipated in R_1, P_1. P_1 = U_1 I P_1 = 4.5\,V × 1.5\,A P_1 = 6.75\,W Step 6: Power dissipated in R_2, P_2. P_2 = U_2 I P_2 =