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

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

ScanSolve AI Answer

Answer

R_1 + R_2 $$

Step 1: Total resistance RR.

R=R1+R2R = R_1 + R_2

R=3Ω+5ΩR = 3\,\Omega + 5\,\Omega

R=8ΩR = 8\,\Omega

Step 2: Total current II.

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

I=12V8ΩI = \frac{12\,V}{8\,\Omega}

I=1.5AI = 1.5\,A

Step 3: Voltage across R1R_1, U1U_1.

U1=IR1U_1 = I R_1

U1=1.5A×3ΩU_1 = 1.5\,A \times 3\,\Omega

U1=4.5VU_1 = 4.5\,V

Step 4: Voltage across R2R_2, U2U_2.

U2=IR2U_2 = I R_2

U2=1.5A×5ΩU_2 = 1.5\,A \times 5\,\Omega

U2=7.5VU_2 = 7.5\,V

Step 5: Power dissipated in R1R_1, P1P_1.

P1=U1IP_1 = U_1 I

P1=4.5V×1.5AP_1 = 4.5\,V \times 1.5\,A

P1=6.75WP_1 = 6.75\,W

Step 6: Power dissipated in R2R_2, P2P_2.

P2=U2IP_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.

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

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 Answer
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 =