The circuit shows a battery of negligible internal resistance connected to three resistors. (a) Calculate the potential difference across the 15 resistor. (b) Calculate the current I1 in the 4.0 resistor. (c) Calculate the current I2 and the resistance R.

Physics
The circuit shows a battery of negligible internal resistance connected to three resistors. (a) Calculate the potential difference across the 15 resistor. (b) Calculate the current I1 in the 4.0 resistor. (c) Calculate the current I2 and the resistance R.

ScanSolve AI Answer

Answer

3.6 V

Sawa Jozee, hebu tusuluhishe tena.

Swali la 1:

a) Kokotoa tofauti ya uwezo (potential difference) kwenye kipingamizi cha 15Ω15 \, \Omega.

Step 1: Tumia Sheria ya Ohm, V=IRV = IR. Mkondo wa umeme unaopita kwenye kipingamizi cha 15Ω15 \, \Omega ni 0.24A0.24 \, \text{A}. V=0.24A×15ΩV = 0.24 \, A \times 15 \, \Omega V=3.6VV = 3.6 \, \text{V} Tofauti ya uwezo kwenye kipingamizi cha 15Ω15 \, \Omega ni 3.6V\boxed{3.6 V}.

b) Kokotoa mkondo wa umeme I1I_1 kwenye kipingamizi cha 4.0Ω4.0 \, \Omega.

Step 1: Tofauti ya uwezo kwenye vipingamizi vilivyounganishwa sambamba (RR na 15Ω15 \, \Omega) ni sawa na 3.6V3.6 \, \text{V} (kutoka sehemu a). Jumla ya tofauti ya uwezo kutoka betri ni 6.0V6.0 \, \text{V}. Tofauti ya uwezo kwenye kipingamizi cha 4.0Ω4.0 \, \Omega ni V4.0ΩV_{4.0\Omega}. Kwa mzunguko wa mfululizo, Vtotal=V4.0Ω+VparallelV_{total} = V_{4.0\Omega} + V_{parallel}. 6.0V=V4.0Ω+3.6V6.0 \, V = V_{4.0\Omega} + 3.6 \, \text{V} V4.0Ω=6.0V3.6VV_{4.0\Omega} = 6.0 \, V - 3.6 \, \text{V} V4.0Ω=2.4VV_{4.0\Omega} = 2.4 \, \text{V}

Step 2: Tumia Sheria ya Ohm kukokotoa I1I_1. I1=V4.0ΩR4.0ΩI_1 = \frac{V_{4.0\Omega}}{R_{4.0\Omega}} I1=2.4V4.0ΩI_1 = \frac{2.4 \, V}{4.0 \, \Omega} I1=0.6AI_1 = 0.6 \, \text{A} Mkondo wa umeme I1I_1 ni 0.6A\boxed{0.6 A}.

c) Kokotoa mkondo wa umeme I2I_2 na kipingamizi RR.

Step 1: Kokotoa mkondo wa umeme I2I_2 kwa kutumia Sheria ya Kirchhoff ya Mikondo. Mkondo wa umeme I1I_1 hugawanyika kuwa I2I_2 na 0.24A0.24 \, \text{A}. I1=I2+0.24AI_1 = I_2 + 0.24 \, \text{A} 0.6A=I2+0.24A0.6 \, A = I_2 + 0.24 \, \text{A} I2=0.6A0.24AI_2 = 0.6 \, A - 0.24 \, \text{A} I2=0.36AI_2 = 0.36 \, \text{A} Mkondo wa umeme I2I_2 ni 0.36A\boxed{0.36 A}.

Step 2: Kokotoa kipingamizi RR. Kipingamizi RR kimeunganishwa sambamba na kipingamizi cha 15Ω15 \, \Omega, hivyo tofauti ya uwezo kwenye RR ni 3.6V3.6 \, \text{V}. Tumia Sheria ya Ohm: R=VI2R = \frac{V}{I_2}. R=3.6V0.36AR = \frac{3.6 \, V}{0.36 \, A} R=10ΩR = 10 \, \Omega Kipingamizi RR ni 10Ω\boxed{10 \Omega}.

That's 2 down. 3 left today — send the next one.

Was this helpful?

Need help with your own homework?

Get instant step-by-step solutions to any question. Free to start.

Ask Your Question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp

Still have questions?

Quick Answer

Sawa Jozee, hebu tusuluhishe tena. Swali la 1: a) Kokotoa tofauti ya uwezo (potential difference) kwenye kipingamizi cha 15 \, .

The circuit shows a battery of negligible internal resistance connected to three resistors. (a) Calculate the potential difference across the 15 resistor. (b) Calculate the current I1 in the 4.0 resistor. (c) Calculate the current I2 and the resistance R.
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
Sawa Jozee, hebu tusuluhishe tena. Swali la 1: a) Kokotoa tofauti ya uwezo (potential difference) kwenye kipingamizi cha 15 \, . Step 1: Tumia Sheria ya Ohm, V = IR. Mkondo wa umeme unaopita kwenye kipingamizi cha 15 \, ni 0.24 \, A. V = 0.24 \, A × 15 \, V = 3.6 \, V Tofauti ya uwezo kwenye kipingamizi cha 15 \, ni 3.6 V. b) Kokotoa mkondo wa umeme I_1 kwenye kipingamizi cha 4.0 \, . Step 1: Tofauti ya uwezo kwenye vipingamizi vilivyounganishwa sambamba (R na 15 \, ) ni sawa na 3.6 \, V (kutoka sehemu a). Jumla ya tofauti ya uwezo kutoka betri ni 6.0 \, V. Tofauti ya uwezo kwenye kipingamizi cha 4.0 \, ni V_4.0. Kwa mzunguko wa mfululizo, V_total = V_4.0 + V_parallel. 6.0 \, V = V_4.0 + 3.6 \, V V_4.0 = 6.0 \, V - 3.6 \, V V_4.0 = 2.4 \, V Step 2: Tumia Sheria ya Ohm kukokotoa I_1. I_1 = V_4.0R_4.0 I_1 = 2.4 \, V4.0 \, I_1 = 0.6 \, A Mkondo wa umeme I_1 ni 0.6 A. c) Kokotoa mkondo wa umeme I_2 na kipingamizi R. Step 1: Kokotoa mkondo wa umeme I_2 kwa kutumia Sheria ya Kirchhoff ya Mikondo. Mkondo wa umeme I_1 hugawanyika kuwa I_2 na 0.24 \, A. I_1 = I_2 + 0.24 \, A 0.6 \, A = I_2 + 0.24 \, A I_2 = 0.6 \, A - 0.24 \, A I_2 = 0.36 \, A Mkondo wa umeme I_2 ni 0.36 A. Step 2: Kokotoa kipingamizi R. Kipingamizi R kimeunganishwa sambamba na kipingamizi cha 15 \, , hivyo tofauti ya uwezo kwenye R ni 3.6 \, V. Tumia Sheria ya Ohm: R = (V)/(I_2). R = 3.6 \, V0.36 \, A R = 10 \, Kipingamizi R ni 10 . That's 2 down. 3 left today — send the next one.