Voltage (Potential) Dividers A voltage (potential) divider is simply 2 resistors connected across a power supply, as shown: 8959 *23"&&$LB E$$$$$$$LB$$$$$$$$$$$$$7$$$$A& L Z$LB +6V R1 + - $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ C $$$$$$$$$$$$$$$$$$$$$$$$$$$E$$$$$$$$[$$$$$$$$7$$$$$P$A 4Ω $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\$Z$[ 6V $$$$$$$$$$$$$$$$$$$$$$$$$$$E$$$$$[$$$$$$7$$$$PA$$$$$$E$$$$$\$A R2 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$CB 8Ω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Ω R1 10 kΩ Ω R1 5 MΩ Ω R2 400 Ω R2 110 kΩ Ω R2 10 MΩ Ω 0V 0V 0V Wheatstone Bridge Circuits A Wheatstone Bridge circuit consists of 2 voltage (potential) dividers connected to a power supply with a voltmeter or sensitive ammeter connected between them, as shown: R1 + + V - V R4 R2 R3 R1 R3 R2 R4 C?$]*+*12)5$>.)*'&'"1) ]3056) If the voltmeter reads 0 V, we say the Wheatstone bridge is balanced. This only happens when: LC LS E LB L\ The voltage of the battery voltage has no effect on this ratio - If you change the battery voltage, the reading on the voltmeter will still be 0 V. NH802*+$O3"(+)% U/#(Z/#7%,%1+# 3&'*.#($'&$0'%(,/14+(',(37"7+$#*(:%/#(@1"%2#%#&(&#7*,(I(Q;8(>#%#&2'+# %/#(&#,',%7+$#(1)(L\8 >.)*'&'"1) (3056)$0&$(*+*12)5D(,1? R1 + - 800 Ω 40 Ω R3 ?Ω R4 LC LS LB E L\ V 6V R2 600 Ω [QQ \Q E PQQ L\ ((((((C&1,,K20"%'-"6'+.D([QQ$7 L\ E$PQQ$7 \Q $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$,1D(L\ E$$PQQ$7 \Q$$E$$SQ$Ω Ω $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$[QQ B?$X1(*+*12)5$>.)*'&'"1) ]3056) If the voltmeter does not read 0 V, we say the Wheatstone bridge is unbalanced. LC LB 5")&$1"'$)M#*+ LS L\ NH802*+$O3"(+)% U/#(Z/#7%,%1+# 3&'*.#($'&$0'%(,/14+(',(0+37"7+$#*8(>#%#&2'+#(%/#(&#7*'+.(1+(%/#(@1"%2#%#&? 8959 *23"&&$LB E$$$$LB$$$$$$$$$$$$$7$$$$A& R1 + - 9Ω 3Ω 8V L Z$LB $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ C R3 $$$$$$$$$$$$$$$$$$$$$$$$$$$$E$$$$$$$$$S$$$$$$$$7$$$$$[$A V R2 3Ω $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$S$Z$S 3Ω R4 $$$$$$$$$$$$$$$$$$$$$$$$$$$$E$$$$$S$$$$$$7$$$$[$A$$$$$$E$$$$$\$A $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$P 8959 *23"&&$L\ E$$$$$$$L\$$$$$$$$$$$$$7$$$$A& L Z$L\ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ S $$$$$$$$$$$$$$$$$$$$$$$$$$$E$$$$$$S$$$$$$$$7$$$$$[$A $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$R$Z$S $$$$$$$$$$$$$$$$$$$$$$$$$$$E$$$$$S$$$$$$7$$$$[$A$$$$$$E$$$$$B$A $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$CB &"/$8959 *23"&&$:"+'%)')3$=:"+'%)')3$3)*5016?$E$$$\$A$J$B$A$E$$$B$A *-HI:H-A=%AD=% *-HI:H-A=%AD=%O-H:=% AD=%O-H:=%BF% O-H:=%BF%AD=% BF%AD=%:.S.BN.% AD=%:.S.BN.%;=C9CA-.I=% :.S.BN.%;=C9CA-.I=%Q ;=C9CA-.I=%Q+]3%9.% 3%9.%=-ID% 9.%=-ID%I-C=M =-ID%I-C=M RX 120 Ω 0V 2V 120 Ω 4 kΩ Ω 0V 2V 9Ω 12 kΩ Ω 10 kΩ Ω 15 kΩ Ω 0V 2V RX 25 kΩ Ω 3.6 kΩ Ω RX #.%=-ID% #.%=-ID%I-C=G% =-ID%I-C=G%I-HI:H-A=% I-C=G%I-HI:H-A=%AD=% I-HI:H-A=%AD=%?BA=.A9-H% AD=%?BA=.A9-H%/9FF=;=.I=% ?BA=.A9-H%/9FF=;=.I=%Q /9FF=;=.I=%Q?E/E3% ?E/E3%-I;BCC% 3%-I;BCC%JUG -I;BCC%JUG 9E=EG%AD=% 9E=EG%AD=%OBHA@=A=;% AD=%OBHA@=A=;%;=-/9.<M OBHA@=A=;%;=-/9.<M + 12 V 3 kΩ Ω 9 kΩ Ω A V B 3 kΩ Ω 3 kΩ Ω 5 kΩ Ω 10 kΩ Ω 0V +5V A V 2 kΩ Ω B 8 kΩ Ω 0V + 10 V 3 kΩ Ω 6 kΩ Ω A 2 kΩ Ω 0V V B 4 kΩ Ω +9V Q-3%5=A=;@9.=% 5=A=;@9.=%AD=% AD=%;=-/9.<% AD=%OBHA@=A=;E Q-3%5=A=;@9.=% AD=%;=-/9.<%B.% ;=-/9.<%B.%AD=% B.%AD=% OBHA@=A=;E 9Ω 6Ω QL3%1A-A=% IB:H/%@-S=% @-S=%AB 3%1A-A=%B.=% 1A-A=%B.=%-HA=;-A9B.% B.=%-HA=;-A9B.%KB:% -HA=;-A9B.%KB:%IB:H/% KB:%IB:H/% @-S=%AB AD=%I9;I:9A% AD=%I9;I:9A%ND9ID% I9;I:9A%ND9ID%NB:H/% ND9ID%NB:H/%?;B/:I=% NB:H/%?;B/:I=%-% ?;B/:I=%-%^_=;B^ -%^_=;B^ ;=-/9.<%B.% ;=-/9.<%B.%AD=% B.%AD=%OBHA@=A=;E% AD=%OBHA@=A=;E% V 6Ω 3Ω 0V Q-3%#/=.A9FK% Q-3%#/=.A9FK%=-ID% #/=.A9FK%=-ID%BF% =-ID%BF%AD=C=% BF%AD=C=%I9;I:9ACG% AD=C=%I9;I:9ACG%:C9.<% I9;I:9ACG%:C9.<%AD=% :C9.<%AD=%FBHHBN9.<% AD=%FBHHBN9.<%H-L=HCM FBHHBN9.<%H-L=HCM (*+*12)5$>.)*'&'"1) (3056)$$$$$$#1(*+*12)5$>.)*'&'"1) (306)$$$$$$1"'$*$>.)*'&'"1) (3056) 10 kΩ Ω 5 kΩ Ω V 2V 5 kΩ Ω 12 kΩ Ω V 2V 10 kΩ Ω 9 kΩ Ω 16 kΩ Ω 10 kΩ Ω V 2V 12 kΩ Ω 7 kΩ Ω 20 kΩ Ω 14 kΩ Ω QL3%6BN% 3%6BN%/B% 6BN%/B%KB:% /B%KB:%S.BN% KB:%S.BN%ND=.% S.BN%ND=.%-%% ND=.%-%%`D=-ACAB.= -%%`D=-ACAB.= U;9/<=%9C% U;9/<=%9C%L-H-.I=/P 9C%L-H-.I=/P aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa QI3%4B;% QI3%4B;%AD=% 4B;%AD=%:.L-H-.I=/% AD=%:.L-H-.I=/%`D=-ACAB.= :.L-H-.I=/%`D=-ACAB.= U;9/<=M Q93%*-HI:H-A=% AD=%OBHA@=A=;b 3%*-HI:H-A=%AD=% *-HI:H-A=%AD=%?BA=.A9-H% AD=%?BA=.A9-H%/9FF=;=.I=% ?BA=.A9-H%/9FF=;=.I=%Q /9FF=;=.I=%Q?E/E3% ?E/E3%-I;BCC% 3%-I;BCC%AD=% -I;BCC%AD=% OBHA@=A=;b Q993% AD=%/9;=IA9B.% /9;=IA9B.%BF% 993%1A-A=% 3%1A-A=%AD=% 1A-A=%AD=% /9;=IA9B.%BF%=H=IA;B.% BF%=H=IA;B.%FHBN% =H=IA;B.%FHBN%AD;B:<D% FHBN%AD;B:<D%AD=% AD;B:<D%AD=%OBHA@=A=;E% AD=%OBHA@=A=;E% Practical Application for a Wheatstone Bridge Circuit When a Wheatstone bridge circuit is balanced, the potential difference (p.d.) across the bridge is always __ V. The formula connecting the resistors in the balanced Wheatstone bridge is: = However, when the resistance of any one resistor in the Wheatstone bridge circuit is changed by a small amount, the bridge becomes ______________ and the above relationship no longer applies - An "out of balance" potential difference (p.d.) is created across the bridge. The "out of balance" potential difference is directly proportional to the change in resistance for the one resistor, as shown on the graph below: out of balance p.d./ V 0 change in resistance/ Ω This forms the basis for many sensor devices - For example, a light meter. + 12 V variable resistor 20 kΩ Ω 0V 10 kΩ Ω V LDR In this Wheatstone bridge circuit, the variable resistor is adjusted until the bridge is balanced - The voltmeter will read ___ V. As the light level changes, the resistance of the LDR will change, so the Wheatstone bridge will now be _________________. An "out of balance" potential difference (p.d.) is created across the bridge, i.e., the voltmeter will show either a positive or negative voltage, depending on whether the light level has increased or decreased. This "out of balance" potential difference is directly proportional to the change in light level. This allows us to calibrate the voltmeter (put a scale on it), so the voltmeter reading will give the actual value for the light level. 1A-A=%ND-A 1A-A=%ND-A /=O9I=%AD9C /=O9I=%AD9C `D=-ACAB.= L;9/<=%I9;I:9A% L;9/<=%I9;I:9A%N9HH I9;I:9A%N9HH L=%A:;.=/% A:;.=/%9.AB% L=% A:;.=/%9.AB%9F 9.AB%9F AD=%)5+ )5+ 9C AD=% ;=?H-I=/%N9AD%-M% Q-3%%AD=;@9CAB; AD=;@9CAB;M% AD=;@9CAB;M%aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa QL3%?;=CC:;=%C=.C9A9O=%;=C9CAB;M%aaaaaaaaaaaaaaaaaaaaaaaaaaaa QI3%CA;-9.%C=.C9A9O=%;=C9CAB;M%aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa Power The power of a circuit component (such as a resistor) tells us how much electrical potential energy the component transforms (changes into other forms of energy) every second: P= E t The following formulae are used to calculate power (P): $$$$$$$$$$$$$$$O$E$AW$$$$$$$$$$$$$O$E$WBL$$$$$$$$$$$$$$$$$O$E$AB L In each case, calculate the power of the resistor: 6V 2A 4Ω 5V 2.5 Ω 1.5 A