Kirchhoff's Laws - Assignment Example

Kirchhoffs Laws Kirchhoffs Laws The values of current and resistance in the circuit. What are these values? R1 Rx R2 R3 Total E(Volts)99999I (Amps)0.9m0.1m4.5m9m14.5mR (Ohms)10k0.2k2k1k14kAs the current flows through the circuit, across the negative (-) cell terminal, some of the current splits to R2 and Rx. Each resistor is rated maximum of 250 mW. If the circuit allows the value to be exceeded, too much heating will be generated, and this situation may lead to other effects. 2. Increase in the values of R2 and Rx and their equal values effects.

What big change occurs in the circuit now? Give a listing of resistance and current values and compare them to the default circuitR1RxR2R3TotalE Volts)99999I (Amps)0.9m4.5m4.5m9m14.5mR (Ohms)10k0.4k4k1k15k Explanation:Current through the smaller resistance between R2 and Rx is less compared to the current through R1 and R3. The current through R1 and R2 is the same as the current through IAmm, R2 and Rx.R1RxR2R3TotalE Volts)99999I (Amps)1.8m4.5m9m9m24.30mR (Ohms)5k0.2k1k1k9kBy lowering the amount of resistance of R1 and R2 in the circuit, the amount of current through the circuit is also increased.

The power is given by P=VI, hence the power dissipated by the default resistor was 0.9watts and hence resistor will not over heat.3. The results of an experiment with an input voltage of 5.0V. What would the new value of the power dissipated by the resistors be in this case? The expected overheat on the resistors especially if they were rated at ¼ watt? Results and ExplanationR1RxR2R3TotalE Volts)55555I (Amps)2.5m5m2.5m5m8mR Ohms)10k0.2k2k1k15kThe power dissipated will be 0.5 watts; the resistor will not be overheated.

R2/R1=Rx/R3Rx=R2/R1xR3 Using Kirchhoff’s first rule to find the currents in junctions A and C A CIx=5m, Ig=0V, I3-Ix+Ig=0(5-5+0)=0Then using Kirchhoff’s second law to find the voltage in the loops ADB and DBC B B A D C DI1-I2-Ig=0Vg= Rx/(R3+Rx) –R2/(R1+R2) Vs = (0.167- 0.167) 5 = 0VR1RxR2R3TotalE (Volts)7.57.57.57.57.5I (Amps)0.00075m0.0375m0.00375m0.0075m0.0005mR (Ohms)10k0.2k2k1k15kUsing Kirchhoff’s first rule to find the currents in junctions A and CIx = 0.00375 Ig =0.

00375I3-Ix+Ig=00.0075-0.00375-0.00375=0Then using Kirchhoff’s second law (Marwedel, 2011) to find the voltage in the loops DBC and ADBVg= Rx/(R3+Rx) –R2/(R1+R2) Vs = (0.167-0.167) 7.5 = 0VReferencesMarwedel, P. (2011). Embedded system design: Embedded systems foundations of cyber- physical systems. Dordrecht: Springer.